Chapter 5.4 Summation

In this section, we talk about a special kind of recursive sequences called series. We will introduce ways to determine if these series converge.

Infinite Series and Their Convergence

For any sequence \((a_n)\), define the sequence \((s_n)\) recursively by

\[ \begin{cases} s_1=a_1\\ s_{n+1}=s_n+a_{n+1} &\text{if }n\in\mathbb{N}. \end{cases} \]

As an example, if \((a_n)\) is a sequence given by \[a_n=\frac{n}{n+1},\] then here are the first four terms of the sequence \((s_n)\):

  • \(s_1=a_1=\frac{1}{2}\)
  • \(s_2=s_1+a_2=\frac{1}{2}+\frac{2}{3}=\frac{7}{6}\)
  • \(s_3=s_2+a_3=\frac{7}{6}+\frac{3}{4}=\frac{23}{12}\)
  • \(s_4=s_3+a_4=\frac{23}{12}+\frac{4}{5}=\frac{163}{60}.\)

The sequence \((s_n)\) is a series, also called the sequence of partial sums of \((a_n).\)

For each \(n\), we call \(s_n\) a finite sum or finite series because \(s_n\) is just a sum of a finite number of terms, which we denote like this:

\[s_n=\sum_{k=1}^n a_k.\]

View the infinite series \(\sum\limits _{k=1}^{\infty}a_k\) formally as a notation for the sequence of partial sums \((s_n)\).

Our goal is to evaluate an infinite series, which means to determine the limit of the sequence of partial sums of the infinite series. That is

The first thing we will do is become familar with the notation with this example.

Example 1

For any integer \(m\), take \(S(m)\) to be given by \[S(m)=\sum_{k=m}^\infty \frac{\sin(k)}{k^2+3}.\] For each of these choices of \(m\), identify a recursive formula for \(S(m)\) so that its sequence of patial sums \((s_n)\) is indexed by natural numbers:

  1. \(m=0\)
  2. \(m=1\)
  3. \(m=5\).

  1. The infinite series is \(S(0)=\sum\limits_{k=0}^{\infty}\frac{\sin(k)}{k^2+3}.\) Here is a recurisve formula for \(S(0)\) that is indexed by the natural numbers. \[ \begin{cases} T(1)=\frac{\sin(0)}{0^2+3}=0\\ T(n+1)=T(n)+\frac{\sin(n)}{n^2+3} \text{if }n\in\mathbb{N}. \end{cases} \]

  2. The infinite series is \(S(1)=\sum\limits_{k=1}^{\infty}\frac{\sin(k)}{k^2+3}.\) Here is a recurisve formula for \(S(1)\) that is indexed by the natural numbers. \[ \begin{cases} T(1)=\frac{\sin(1)}{1^2+3}=\frac{\sin(1)}{4}\\ T(n+1)=T(n)+\frac{\sin(n+1)}{(n+1)^2+3} \text{if }n\in\mathbb{N}. \end{cases} \]

  3. The infinite series is \(S(5)=\sum\limits_{k=5}^{\infty}\frac{\sin(k)}{k^2+3}.\) Here is a recurisve formula for \(S(5)\) that is indexed by the natural numbers. \[ \begin{cases} T(1)=\frac{\sin(5)}{5^2+3}=\frac{\sin(5)}{28}\\ T(n+1)=T(n)+\frac{\sin(n+5)}{(n+5)^2+3} \text{if }n\in\mathbb{N}. \end{cases} \]

Determining what an infinite series equals can be done by determining a simple formula for its partial sums. In such cases, the infinite series equals the limit of this simple formula.

We will investigate three kinds of series that have such nice formulas.

The first one is called an arithmetic series, which comes from a sequence that is of arithmetic progression.

A sequence \((a_n)\) is an arithmetic progression if there are real numbers \(A\) and \(B\) so that for each \(n\),

\[a_n=An+B.\]

The sequence of partial sums of an arithmetic progression \((a_n)\) is an arithmetic series.

For these type of series, there is a general formula for its partial sums.

First, for any natural number \(n\), we have

\[\begin{align*} 2(1+2+3+4+\dots+n)&=1+2+3+\dots+(n-3)+(n-2)+(n-1)+n+n+(n-1)+(n-2)+\dots+3+2+1\\ &=(n-1+1)+(n-1+1)+(n-2+2)+(n-2+2)+(n-3+3)+(n-3+3)+\dots+n+n\\ &=\underbrace{n+n+n+n+n+n\dots+n+n}_{n+1\text{ of these}}\\ &=n(n+1), \end{align*}\]

and so

\[1+2+3+4+\dots+n=\frac{n(n+1)}{2}\]

which can be denoted as

\[\sum_{k=1}^n k=\frac{n(n+1)}{2}.\]

The commutativity and distributivity of finite sums and the formula for the sums of the simple arithmetic series implies that

\[\begin{align*} \sum_{k=1}^n(Ak+B)&=\sum_{k=1}^n Ak+\sum_{k=1}^n B\\ &=A\sum_{k=1}^n k+\sum_{k=1}^n B\\ &=A\frac{n(n+1)}{2} k+Bn. \end{align*}\]

This is the general formula for an arithmetic series.

Example 2

Evalaute these finite series:

  1. \(\sum\limits_{k=1}^{120}(7k-9)\)

  2. \(\sum\limits_{k=205}^{920}(5k-2)\)

  1. Use the formula to get \[\begin{align*} \sum_{k=1}^{120}(7k-9)&=7\sum_{k=1}^{120}k-9\sum_{k=1}^{120}1\\ &=7\cdot\frac{120\cdot 121}{2}-9\cdot 120. \end{align*}\]

  2. Rewrite the series as \(\sum\limits_{k=205}^{920}(5k-2)=\sum\limits_{k=1}^{920}(5k-2)-\sum\limits_{k=1}^{205}(5k-2)\) and so \[\begin{align*} \sum_{k=205}^{920}(5k-2)&=\sum_{k=1}^{920}(5k-2)-\sum_{k=1}^{205}(5k-2)\\ &=\left[5\cdot\frac{920\cdot 921}{2}-2\cdot 920\right]-\left[5\cdot\frac{204\cdot 205}{2}+2\cdot 204\right]\\ &=\frac{5}{2}(920\cdot 921-204\cdot 205)-2(920-204). \end{align*}\]

The other kind of series is a telescoping series.

A telescoping series is a series of the form \(\sum\limits_{k=1}^n(f(k)-f(k+1)).\)

Expand the sum and regroup the terms to obtain the following formula:

\[\begin{align*} \sum_{k=1}^n (f(k)-f(k+1))&=(f(1)-f(2))+(f(2)-f(3))+(f(3)-f(4))+\dots+(f(n-2)-f(n-1))+(f(n-1)-f(n))\\ &=f(1)+(f(2)-f(2))+(f(3)-f(3))+(f(4)-f(4))+\dots+(f(n-2)-f(n-2))+(f(n-1)-f(n-1))-f(n)\\ &=f(1)-f(n). \end{align*}\]

Sometimes, the sum is easily recognizable as a telescoping series. Other times, we need to rewrite the terms. Practice this with the following example.

Example 3

Evaluate these finite series:

  1. \(\sum\limits_{k=1}^{100}\left(\frac{1}{k}-\frac{1}{k+1}\right)\)

  2. \(\sum\limits_{k=1}^{100}\frac{1}{k(k+1)}\)

  1. This is a telescoping series as the summand is of the form \(f(k)-f(k+1)\), where \(f(k)=\frac{1}{k}\). And so \[\begin{align*} \sum_{k=1}^{100}\left(\frac{1}{k}-\frac{1}{k+1}\right)&=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{3}\right)+\dots+\left(\frac{1}{98}-\frac{1}{99}\right)+\left(\frac{1}{99}-\frac{1}{100}\right)+\left(\frac{1}{100}-\frac{1}{101}\right)\\ &=1+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{98}-\frac{1}{98}\right)+\left(\frac{1}{99}-\frac{1}{99}\right)+\left(\frac{1}{100}-\frac{1}{100}\right)-\frac{1}{101}\\ &=1-\frac{1}{101}\\ &=\frac{100}{101}. \end{align*}\]

  2. Rewrite the summand \(\frac{1}{k(k+1)}\) like this: \[\frac{1}{k(k+1)}=\frac{A}{k}+\frac{B}{k+1}\] where \(A\) and \(B\) are constants to be determined. Combine the right side to get \[\frac{A}{k}+\frac{B}{k+1}=\frac{A(k+1)+Bk}{k(k+1)}=\frac{(A+B)k+A}{k(k+1)}.\] So, \[\frac{1}{k(k+1)}=\frac{A}{k}+\frac{B}{k+1}\quad\text{is equivalent to}\quad \frac{1}{k(k+1)}=\frac{(A+B)k+A}{k(k+1)}.\] The equality holds if and only if the numerator and denominator are equal. This means \[1=(A+B)k+A,\] which is equivalent to the following system: \[ \begin{cases} 0=A+B\\ 1=A. \end{cases} \] Thus \(A=1\) and \(B=-1\). The summand then looks like this \[\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}.\] The summand is now exactly the same as the summand from part a, so \[\sum_{k=1}^{100}\frac{1}{k(k+1)}=\frac{100}{101}.\]

In this next example, practice finding the decomposition so that the sum can be evaluated. Also, write out the terms to get an idea on how the cancellation works.

Example 4

Evalute these finite series:

  1. \(\sum\limits_{k=1}^{100}\left(\frac{1}{k-1}-\frac{1}{k+3}\right)\)

  2. \(\sum\limits_{k=1}^{100}\frac{1}{k(k+1)}\)

  1. The series is \[\begin{align*}\sum\limits_{k=1}^{100}\left(\frac{1}{k-1}-\frac{1}{k+3}\right)&=1-\frac{1}{5}+\frac{1}{2}-\frac{1}{6}+\frac{1}{3}-\frac{1}{7}+\frac{1}{4}-\frac{1}{8}+\frac{1}{5}-\frac{1}{9}+\dots+\frac{1}{95}-\frac{1}{99}+\frac{1}{96}-\frac{1}{100}+\frac{1}{97}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{99}-\frac{1}{103}\\ &=1-\cancel{\frac{1}{5}}+\frac{1}{2}-\cancel{\frac{1}{6}}+\frac{1}{3}-\cancel{\frac{1}{7}}+\frac{1}{4}-\cancel{\frac{1}{8}}+\cancel{\frac{1}{5}}-\cancel{\frac{1}{9}}+\dots+\cancel{\frac{1}{95}}-\cancel{\frac{1}{99}}+\cancel{\frac{1}{96}}-\frac{1}{100}+\cancel{\frac{1}{97}}-\frac{1}{101}+\cancel{\frac{1}{98}}-\frac{1}{102}+\cancel{\frac{1}{99}}-\frac{1}{103}\\ &=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{100}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}.\end{align*}\]
  2. Rewrite the summand \(\frac{1}{k^2+2k-3}\) like this: \[\frac{1}{k^2+2k-3}=\frac{1}{(k-1)(k+3)}=\frac{A}{k-1}+\frac{B}{k+3}\] where \(A\) and \(B\) are constants to be determined. Combine the right side to get \[\frac{A}{k-1}+\frac{B}{k+3}=\frac{A(k+3)+B(k-1)}{(k-1)(k+3)}=\frac{(A+B)k+3A-B}{k(k+1)}.\] So, \[\frac{1}{k^2+2k-3}=\frac{A}{k-1}+\frac{B}{k+3}\quad\text{is equivalent to}\quad \frac{1}{(k-1)(k+3)}=\frac{(A+B)k+3A-B}{k(k+1)}.\] The equality holds if and only if the numerator and denominator are equal. This means \[1=(A+B)k+3A-B,\] which is equivalent to the following system: \[ \begin{cases} 0=A+B\\ 1=3A-B. \end{cases} \] Thus \(A=\frac{1}{4}\) and \(B=-\frac{1}{4}\). The summand then looks like this \[\frac{1}{k^2+2k-3}=\frac{1}{4}\left(\frac{1}{k}-\frac{1}{k+3}\right).\] The summand is now exactly the same as the summand from part a with a scaling of \(\frac{1}{4}\), so \[\sum_{k=2}^{100}\frac{1}{k^2+2k-3}=\frac{1}{4}\sum_{k=2}^{100}\left(\frac{1}{k-1}-\frac{1}{k+3}\right)=\frac{1}{4}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{100}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}\right).\]

The other kind of sequence is called a geometric progression. A sequence \((a_n)\) is a geometric progression if there are nonzero real numbers \(A\) and \(r\) so that for each \(n\), \[a_n=Ar^n.\] The sequence of partial sums of \((a_n)\) is a geometric series. This is an example of a geometric series: \[\sum_{n=1}^\infty 2\left(\frac{3}{5}\right)^n.\] The formula for its partial sums can be obtained by noticing that for any real number \(r\) not equal to \(1\) and natural number \(n\), the equality

\[(1-r)(1+r+r^2+\dots+r^n)=1-r^{n+1}\] implies that \[1+r+r^2+\dots+r^n=\frac{1-r^{n+1}}{1-r}.\]

Hence, \[\sum_{n=0} r^n=\frac{1-r^{n+1}}{1-r}.\]

Example 5

Evalute these finite series:

  1. \(\sum\limits_{k=0}^{100}\left(\tfrac{1}{3}\right)^k\)

  2. \(\sum\limits_{k=1}^{200}5^k\)

  3. \(\sum\limits_{k=0}^{50}\cos^{2k}(\theta)\)

  1. The series equals \[\begin{align*} \sum_{k=0}^{100}\left(\frac{1}{3}\right)^k&=1+\frac{1}{3}+\left(\frac{1}{3}\right)^2+\dots+\left(\frac{1}{3}\right)^{100}\\ &=\frac{1-\left(\frac{1}{3}\right)^{101}}{1-\frac{1}{3}}\\ &=\frac{3}{2}\cdot \left(1-\left(\frac{1}{3}\right)^{101}\right). \end{align*}\]

  2. Rewrite the series like this: \[\begin{align*} \sum_{k=1}^{200}5^k&=5^4+5^5+5^6+\dots+5^{200}\\ &=5^4\left(1+5+5^2+\dots+5^{196}\right). \end{align*}\] Use the formula on \(1+5+5^2+\dots+5^{196}\) to get \[\begin{align*} \sum_{k=1}^{200}5^k&=5^4\left(1+5+5^2+\dots+5^{196}\right)\\ &=5^4\cdot \frac{1-5^{197}}{1-5}\\ &=5^4\cdot \frac{5^{197}-1}{4}. \end{align*}\]

  3. When \(\theta=\pi m\) and \(m\) is an integer, \(\cos^2(\theta)=1\) and so \[\begin{align*} \sum\limits_{k=0}^{50}\cos^{2k}(\theta)&&=\underbrace{1+1+\dots+1}_{51\text{-times}}\\ &=51. \end{align*}\] When \(\theta\not=\pi m\) where \(m\) is an integer, the series is \[\begin{align*} \sum\limits_{k=0}^{50}\cos^{2k}(\theta)&=1+\left(\cos^2(\theta)\right)+\left(\cos^2(\theta)\right)^2+\left(\cos^2(\theta)\right)^3+\dots+\left(\cos^2(\theta)\right)^{50}\\ &=\frac{1-(\cos^2(\theta))^{51}}{1-\cos^2(\theta)}. \end{align*}\]

Therefore, \[ \sum_{k=0}^{50}\cos^{2k}(\theta)= \begin{cases} \frac{1-(\cos^2(\theta))^{51}}{1-\cos^2(\theta)}&&\text{if }\theta\not=\pi m\text{ and }m\text{ is an integer}\\ 51&&\text{if }\theta=\pi m\text{ and }m\text{ is an integer} \end{cases} \]

Infinite series appear in problems that involve infinite self-similar decomposition.

For example, take a straight ruler of length \(2\):

Divide this ruler into two pieces of length \(1\):

Subdivide the left half of the ruler into two pieces of length \(\frac{1}{2}\):

Continue this process of subdivision of the left half of a segment of the subdivided ruler to obtain this decomposition of a ruler

The length of the ruler should be the sum of the lengths of all of its components.

The lengths of these components should be \(1\),\(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\),\(\frac{1}{16}\) and so on.

The total length of the ruler is \(2\), which means that it should be the case that

\[1+\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+\dots+\frac{1}{2^n}+\dots=2.\]

What does the left-hand side of the above equality mean? View it is a limit.

An infinite series is a procedure for summing infinitely many real values.

Denote \(\sum\limits_{k=1}^{\infty}a_k\) or by \(\sum a_n\) the limit

\[\sum_{k=1}^{\infty}a_k=\lim_{n\to\infty}\sum_{k=1}^{n}a_k.\]

When this limit exists, we call \((a_n)\) summable or that the infinite series \(\sum a_n\) is convergent.

An important thing to note is that if \(|r|<1\), then \(\lim\limits_{n\to\infty}r^n=0.\) However, if \(|r|>1\), then \(r^{n}\) diverges to either \(\infty\) or \(-\infty\). Use these facts to determine the following.

Example 6

Use both the limit of the partial sums and the given decomposition of the equilateral triangle to determine this infinite series: \[\sum_{k=1}^{\infty}\frac{1}{4^k}.\]

By the using the picture, we see the area the triangle \(T_1\) is \(\frac{1}{4}\) of the total area. The area of triangle \(T_2\) is \(\frac{1}{4}\) of the area of \(T_1\). The area of triangle \(T_3\) is \(\frac{1}{4}\) of the area of \(T_2\). In general, the area of triangle \(T_{k+1}\) is \(\frac{1}{4}\) of the area of \(T_k\). The area of \(T_k\) is \(\frac{1}{3}\) the area of the \(k^{\text{th}}\) level. So \[\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\dots=\frac{1}{3}.\] Now, calculate the sum by evaluating the limit: \[\begin{align*} \sum_{k=1}^{\infty}\frac{1}{4^k}&=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{4^k}\\ &=\frac{1}{4}\lim_{n\to\infty}\sum_{k=0}^n\frac{1}{4^k}\\ &=\frac{1}{4}\lim_{n\to\infty}\frac{1-\left(\frac{1}{4}\right)^{n+1}}{1-\frac{1}{4}}\\ &=\frac{1}{4}\frac{1-0}{1-\frac{1}{4}}\\ &=\frac{1}{3}. \end{align*}\]

In order to evaluate more complicated sums, it is important to establish some properties.

Take \((a_n)\) and \((b_n)\) to both be summable sequences. For any real numbers \(C_1\) and \(C_2\),

\[\sum(C_1a_n+C_2b_n)=C_1\sum a_n+C_2\sum b_n.\]

Example 7

For any sequences \((a_n)\) and \((b_n)\) with \[\sum a_n=7\quad\text{and}\quad\sum b_n=2,\] evaluate \(\sum c_n\) for each of these choices of sequence \((c_n)\):

  1. \(c_n=-3a_n\)
  2. \(c_n=a_n+b_n\)
  3. \(c_n=2a_n-5b_n\)
  1. Use the sum properties to obtain: \[\sum -3a_n=-3\sum a_n=-3(7)=-21.\]
  2. Use the sum properties to obtain: \[\sum (a_n+b_n)=\sum a_n+\sum b_n=7+2=9.\]
  3. Use the sum properties to obtain: \[\sum (2a_n-5b_n)=2\sum a_n-5\sum b_n=2\cdot 7-5\cdot 2=4.\]

It is very difficult to work with series because many things can go wrong when manipulating them. For example, consider the following manipulation:

\[\begin{align*} \frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\dots\right)&=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\dots\right)-\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)\\ &=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\dots. \end{align*}\]

Multiply the left-hand side to get

\[\begin{align*} \frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\dots\right)&=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\dots, \end{align*}\]

which means

\[\begin{align*} \frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\dots\right)&=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\dots\\ \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\dots&=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\dots. \end{align*}\]

Subtract the left-hand side from both sides to get

\[\begin{align*} 0&=\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\dots\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\dots\right)\\ &=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\dots. \end{align*}\] However, this is impossible because series is a sum of positive summands.

In fact, the series \(\sum\limits_{k=1}^\infty\frac{1}{k}\), known as the harmonic seies, diverges.

In general, not every series will converge nor will ever series have a closed form for what it converges too. However, there are some that do.

For example, in 1650 Pietro Mengoli challenged his colleagues to find a closed form for this sum:

\[\sum_{n=1}^{\infty}\frac{1}{n^2}.\]

This is known as the Basel problem.

In 1734, a 28 year old Euler proved that

\[\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}.\]

What other series have known formulas?

One of them is geometric series.

For any \(r\) in \((-1,1)\),

\[\sum_{k=0}^{\infty}r^k=\lim_{n\to\infty}\sum_{k=0}^nr^k=\lim_{n\to\infty}\left(\frac{1-r^{n+1}}{1-r}\right)=\frac{1}{1-r},\]

and so

\[\sum_{k=0}^{\infty}r^k=\frac{1}{1-r}.\]

Example 8

Evaluate these series:

  1. \(\sum\limits_{k=3}^{\infty}\left(\tfrac{4}{7}\right)^k\)
  2. \(\sum\limits_{k=0}^{\infty}\left(-\tfrac{1}{3}\right)^k\)
  1. Because \(r=\frac{4}{7}\) is in \((-1,1)\), the series converges. To determine what it converges to, rewrite the series like this: \[\begin{align*} \sum\limits_{k=3}^{\infty}\left(\frac{4}{7}\right)^k&=\left(\frac{4}{7}\right)^3+\left(\frac{4}{7}\right)^4+\left(\frac{4}{7}\right)^5+\dots\\ &=\left(\frac{4}{7}\right)^3\cdot\left(1+\left(\frac{4}{7}\right)+\left(\frac{4}{7}\right)^2+\dots\right). \end{align*}\] The series \(1+\left(\frac{4}{7}\right)+\left(\frac{4}{7}\right)^2+\dots=\sum\limits_{k=0}^{\infty}\left(\frac{4}{7}\right)^k\) converges to \(\frac{1}{1-\frac{4}{7}}\) and so \[\begin{align*} \sum\limits_{k=3}^{\infty}\left(\frac{4}{7}\right)^k&=\left(\frac{4}{7}\right)^3\cdot\left(1+\left(\frac{4}{7}\right)+\left(\frac{4}{7}\right)^2+\dots\right)\\ &=\left(\frac{4}{7}\right)^3\cdot\frac{1}{1-\frac{4}{7}}\\ &=\frac{64}{147}. \end{align*}\]
  2. Because \(r=-\frac{1}{3}\) is in \((-1,1)\), the series converges. It converges to \[\begin{align*} \sum\limits_{k=0}^{\infty}\left(-\frac{1}{3}\right)^k&=1-\frac{1}{3}+\left(-\frac{1}{3}\right)^2+\dots\\ &=\frac{1}{1-\left(-\frac{1}{3}\right)}\\ &=\frac{3}{4}. \end{align*}\]

A telescoping series is another special kind of series that can converge.

For any null sequence \((f(n))\),

\[\sum_{k=m}^{\infty}(f(k)-f(k+1))=\lim_{n\to\infty}\sum_{k=m}^{n}(f(k)-f(k+1))=\lim_{n\to\infty}(f(m)-f(n))=f(m),\]

and so

\[\sum_{k=m}^{\infty}(f(k)-f(k+1))=f(m).\]

Example 9

Evaluate these series:

  1. \(\sum\limits_{n=4}^{\infty}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
  2. \(\sum\limits_{n=2}^{\infty}\frac{1}{(n+1)(n+3)}\)
  1. This is a telescoping series with \((f(n))=\left(\frac{1}{\sqrt{n}}\right)\). This is a null sequence, so the series converges and it converges to \[\begin{align*} \sum\limits_{n=4}^{\infty}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)&=\left(\frac{1}{\sqrt{4}}-\frac{1}{\sqrt{5}}\right)+\left(\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{6}}\right)+\left(\frac{1}{\sqrt{6}}-\frac{1}{\sqrt{7}}\right)+\dots\\ &=\frac{1}{\sqrt{4}}\\ &=\frac{1}{2}. \end{align*}\]
  2. Rewrite the summand \(\frac{1}{(n+1)(n+3)}\) like this: \[\frac{1}{(n+1)(n+3)}=\frac{A}{n+1}+\frac{B}{n+3}\] where \(A\) and \(B\) are constants to be determined. Combine the right side to get \[\frac{A}{n+1}+\frac{B}{n+3}=\frac{A(n+3)+B(n+1)}{(n+1)(n+3)}=\frac{(A+B)n+3A+B}{(n+1)(n+3)}.\] So, \[\frac{1}{(n+1)(n+3)}=\frac{A}{n+1}+\frac{B}{n+3}\quad\text{is equivalent to}\quad \frac{1}{(n+1)(n+3)}=\frac{(A+B)n+3A+B}{(n+1)(n+3)}.\] The equality holds if and only if the numerator and denominator are equal. This means \[1=(A+B)n+3A+B,\] which is equivalent to the following system: \[ \begin{cases} 0=A+B\\ 1=3A+B. \end{cases} \] Thus \(A=\frac{1}{2}\) and \(B=-\frac{1}{2}\). The summand then looks like this \[\frac{1}{(n+1)(n+3)}=\frac{1}{2}\left(\frac{1}{n+1}-\frac{1}{n+3}\right).\] Rewrite the sum in (b) as a telescoping series to obtain the equality \[\begin{align*} \sum_{n=2}^{\infty}\frac{1}{(n+1)(n+3)}&=\frac{1}{2}\sum_{n=2}^{\infty}\left(\frac{1}{n+1}-\frac{1}{n+3}\right)\\ &=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+\frac{1}{5}-\frac{1}{7}+\dots\right)\\ &=\frac{1}{2}\left(\frac{1}{3}-\cancel{\frac{1}{5}}+\frac{1}{4}-\cancel{\frac{1}{6}}+\cancel{\frac{1}{5}}-\cancel{\frac{1}{7}}+\dots\right)\\ &=\frac{1}{2}\left(\frac{1}{3}+\frac{1}{4}\right)\\ &=\frac{7}{24}. \end{align*}\]

Some Convergence Tests

In this subsection, we develop tests we determine if certain series converge or diverge.

First, the terms of a convergent series must be a null sequence. Here is why:

For any sequence \((s_n)\) of partial sums of \(\sum a_n\), if there is an \(s_0\) so that \((s_n)\) converges to \(s_0\), then there is a null sequence \((\varepsilon_n)\) so that for each natural number \(n\), \[s_n=s_0+\varepsilon_n.\] Now consider the sequence \((s_{n+1}-s_n).\) The terms of the sequence look like this: \[|s_{n+1}-s_n|=|a_{n+1}|.\] However, we also that have that \[\begin{align*} |a_{n+1}|&=|s_{n+1}-s_n|\\ &=|s_0+\varepsilon_{n+1}-s_0-\varepsilon_{n}|\\ &=|\varepsilon_{n+1}-\varepsilon_n|\\ &\leq |\varepsilon_{n+1}|+|\varepsilon_n|. \end{align*}\] Therefore, \[0\leq |a_{n+1}|\leq |\varepsilon_{n+1}|+|\varepsilon_n|.\] Since \((|\varepsilon_{n+1}|+|\varepsilon_n|)\) is a null sequence, \((a_{n+1})\) is a null sequence by the sandwich theorem for null sequence.

Therefore, the simplest way to check that a given series \(\sum a_n\) diverges is to verify that \((a_n)\) is not a null sequence.

Example 10

Provide an argument that shows that the series \(\sum \frac{1-n^2}{2n^2+5}\) is divergent.

Determine whether \((a_n)\) is a null sequence or not, where \(a_n=\tfrac{1-n^2}{2n^2+5}\). Calculate the limit to get that

\[ \lim_{n\to\infty}\frac{1-n^2}{2n^2+5}=-\frac{1}{2}. \]

This is not a null sequence, so the series is divergent.

We were able to conclude that the series diverges because its underlining sequence is not a null sequence. However, it is not enough to show a series’ underlining sequence is a null sequence to conclude it converges. In fact, we have a seen a series that diverges despite its underlining sequence diverges: the harmonic series.

We can show that the series diverges because its sequence of partial sums is increasing and unbounded. So in what situations can we conclude a series converges? We can use the Monotone series test.

Monotone Series Test

For any monotone sequence \((a_n)\) with a bounded sequence of partial sums, \(\sum a_n\) is convergent.

Use this test to show that the following sequence is convergent.

Example 11

Use the fact that for each natural number \(n\) that is larger than \(1\),

\[ \frac{1}{n^2}<\frac{1}{n(n-1)}=\frac{1}{n-1}-\frac{1}{n}, \]

to show that \(\sum\frac{1}{k^2}\) is convergent.

The partial sums \((s_n)=\left(\sum\limits_{k=1}^n\tfrac{1}{k^2}\right)\) is an increasing sequence because the summand is positive. Now use the fact given in the Example to get the following upper bound:

\[ \begin{align*} s_n&=1+\sum_{k=2}^n\frac{1}{k^2}\\ &<1+\sum_{k=2}^n\frac{1}{k(k-1)}\\ &=1+\sum_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k}\right)\\ &=1+1\\ &=2. \end{align*} \]

Because \((s_n)\) is a positive, increasing sequence that is bounded above by \(2\), it is convergent.

For some series, it is crucial to be able to manipulate the terms in order to determine what the series converges to. However, as we saw with the harmonic series, manipulating the terms without knowing it converges produced an impossible result. So, we need to be able to identify when we can manipulate the terms to get the right answer. This is where absolutely convergent series come in.

A series \(\sum a_n\) is absolutely convergent if \(\sum |a_n|\) is convergent.

Here are two important facts:

  • Every absolutely convergent series is convergent;
  • For any invertible function \(\sigma\) from \(\mathbb{N}\) to \(\mathbb{N}\), if \(\sum a_n\) is absolutely convergent, then \[\sum_n a_{\sigma(n)}=\sum_n a_n.\]

The series \(\sum a_{\sigma(n)}\) is called a rearrangement of the series \(\sum a_n\). The second bullet point is crucial, because it says that the rearrangement of the terms of an absolutely convergent series will still converge to whatever number the original series converges to.

A series \(\sum a_n\) is conditionally convergent if \(\sum a_n\) is convergent, but \(\sum |a_n|\) is divergent.

To determine which series are absolutely convergent and conditionally convergent, we need additional test to assist us. To start, here is the comparison test.

Comparison Test

For any positive sequence \((a_n)\) and \((b_n)\),

  • if \(\sum a_n\) converges and if for each \(n\), \(b_n\) is less than or equal to \(a_n\), then \(\sum b_n\) converges;
  • if \(\sum a_n\) diverges to \(\infty\) and if for each \(n\), \(b_n\) is greater than \(a_n\), then \(\sum b_n\) diverges to infinity.

Use the comparison test to answer this example. It is crucial in these examples to find estimates that involve series we know converges.

Example 12

Use the comparison test to determine the convergence or divergence of these series:

  1. \(\sum\limits_{k=1}^\infty\frac{1}{k^2+2k-7}\)
  2. \(\sum\limits_{k=1}^{\infty}\frac{1}{\sqrt{k^2+5}}\)
  1. If \(k\geq 4,\) then \(2k-7>0\), and so \[0<\frac{1}{k^2+2k-7}<\frac{1}{k^2}.\] Therefore, \[\sum_{k=4}^{\infty}\frac{1}{k^2+2k-7}<\sum_{k=4}^{\infty}\frac{1}{k^2},\] where the series on the right converges. The comparison test implies that \(\sum\limits_{k=4}^{\infty}\frac{1}{k^2+2k-7}\) is convergent and so the original series is also convergent.
  2. For each \(k\) in \(\mathbb{N}\) with \(k\geq 3\), we have \(k^2>5\), and so \[\frac{1}{\sqrt{k^2+5}}\geq \frac{1}{\sqrt{k^2+k^2}}=\frac{1}{\sqrt{2}}\cdot \frac{1}{k}.\]
    Therefore, \[ \sum_{k=3}^\infty \frac{1}{\sqrt{k^2+5}}\geq \frac{1}{\sqrt{2}}\sum_{k=3}^{\infty}\frac{1}{k},\]
    where the series on the right diverges to \(\infty\) because it is the harmonic series scaled by \(\tfrac{1}{\sqrt{2}}.\) The comparison test implies that \(\sum\limits_{k=3}^{\infty}\frac{1}{\sqrt{k^2+5}}\) is divergent. Since \(\sqrt{k^2+5}\) is positive for all \(k\), \[\sum_{k=1}^{\infty}\frac{1}{\sqrt{k^2+5}}=\frac{1}{\sqrt{6}}+\frac{1}{\sqrt{9}}+\sum_{k=3}^{\infty}\frac{1}{\sqrt{k^2+5}},\]
    and so \(\sum\limits_{k=1}^{\infty}\frac{1}{\sqrt{k^2+5}}\) is divergent.

The next test we can use is called the limit comparison test.

Limit Comparison Test

For any positive sequences \((a_n)\) and \((b_n)\), if there is a positive real number \(L\) so that \(\lim\limits_{n\to\infty}\tfrac{a_n}{b_n}=L,\) then

  • \(\sum a_n\) converges if and only if \(\sum b_n\) converges,
  • \(\sum a_n\) diverges to infinity if and only if \(\sum b_n\) diverges to infinity.

To use the limit comparison test effectively, try to determine the asymptotic behavior of the the sequence in order to identify a sequence to compare it to.

Example 13

Use the limit comparison test to determine the convergence or divergence of these series:

  1. \(\sum\limits_{n=1}^{\infty}\tfrac{2^n+n}{5^n+7}\)

  2. \(\sum\limits_{n=1}^{\infty} n\sin\left(\tfrac{1}{n^3}\right)\)

  1. The terms \(\frac{2^n+n}{5^n+7}\) behavior asymptotically like \(\frac{2^n}{5^n}.\) Take \((a_n)\) and \((b_n)\) to be sequences given by \(a_n=\tfrac{2^n+n}{5^n+7}\) and \(b_n=\tfrac{2^n}{5^n}.\) Rewrite the quotient \(\tfrac{a_n}{b_n}\) like this: \[ \begin{align*} \frac{a_n}{b_n}&=\frac{2^n+n}{5^n+7}\cdot \frac{5^n}{2^n}\\ &=\left(\frac{2^n+n}{2^n}\right)\cdot\left(\frac{5^n}{5^n+7}\right)\\ &=\left(1+\frac{n}{2^n}\right)\cdot \left(\frac{1}{1+\frac{7}{\,5^n}}\right)\ \end{align*} \] The terms \(1+\frac{n}{\,2^n}\) and \(\frac{1}{1+\frac{7}{\,5^n}}\) both tend to \(1\) as \(n\) goes to infinity. Therefore, \[\lim_{n\to\infty}\frac{a_n}{b_n}=1.\] Thus by the limit comparison test, \(\sum a_n\) converges if and only if \(\sum b_n\) converges. The series \(\sum b_n=\sum\left(\tfrac{2}{5}\right)^n\) is a geometric series with \(r=\frac{2}{5}\). Because \(|r|<1\), it converges. Therefore, \(\sum a_n=\sum \tfrac{2^n+n}{5^n+7}\) converges as well.

  2. The terms \(n\sin\left(\tfrac{1}{n^3}\right)\) behaves like \(n\cdot \frac{1}{n^3}=\frac{1}{n^2},\) Take \((a_n)\) and \((b_n)\) to be the sequences given by \(a_n=n\sin\left(\tfrac{1}{n^3}\right)\) and \(b_n=\frac{1}{n^2}.\) Rewrite the quotient \(\frac{a_n}{b_n}\) like this: \[\begin{align*}\frac{a_n}{b_n}&=\frac{n\sin\left(\frac{1}{n^3}\right)}{\frac{1}{n^2}}\\&=\frac{\sin\left(\frac{1}{n^3}\right)}{\frac{1}{n^3}}.\end{align*}\] As \(n\) goes to infinity, \(\tfrac{1}{n^3}\) tends to \(0\) so we have that\(\frac{\sin\left(\frac{1}{n^3}\right)}{\frac{1}{n^3}}\) tends to \(1\). Therefore, \(\lim\limits_{n\to\infty}\tfrac{a_n}{b_n}=1.\) Because \(\sum\tfrac{1}{n^2}\) is convergent we have by the limit comparison test that \(\sum n\left(\frac{1}{n^3}\right)\) also converges.

The tests we have used so far only work for the series \(\sum a_n\) where \(a_n\) is positive. We will extend our analysis to series of the form \(\sum (-1)^n a_n\) where \((a_n)\) is a positive monotone null sequence.

The Alternating Series Test

For any positive monotone null sequence \((a_n)\), \(\sum (-1)^na_n\) is convergent.

The key reason why this is true is because the sequence of partial sums of \(((-1)^na_n)\) , \(s_n\), produces two subsequences. The subsequence \((s_{2n-1})\) is decreasing while the subsequence \((s_{2n})\) is increasing. So, for each natural number \(n\), \(s_{2n-1}\) is greater than \(s_{2n}\) and so \((s_{2n-1}-s_{2n})\) is a null sequence. The nested interval property and the Archimedean property of \(\mathbb{R}\) together imply that \((s_n)\) is convergent.

Use this test to complete the following example.

Example 14

Use the alternating series test to determine the convergence of these series and then determine which of these series are conditionally convergent and which are absolutely convergent:

  1. \(\sum\limits_{k=1}^{\infty}\tfrac{(-1)^k}{k}\)
  2. \(\sum\limits_{k=1}^{\infty}\tfrac{(-1)^k}{3^k}\)
  3. \(\sum\limits_{k=1}^{\infty}\tfrac{(-1)^k}{\sqrt{3k+2}}\)
  4. \(\sum\limits_{k=1}^{\infty}\tfrac{(-1)^k}{k^2+5}\)
  1. Take \((a_k)\) to be the sequence given by \(a_k=\tfrac{1}{k}\). The sequence is decreasing and converges to zero, so by the Alternating Series Test, \(\sum\limits_{k=1}^{\infty}\tfrac{(-1)^k}{k}\) converges. However, \(\sum\limits_{k=1}^{\infty}\left|\tfrac{(-1)^k}{k}\right|=\sum\limits_{k=1}^{\infty}\tfrac{1}{k}\) diverges, so the series \(\sum\limits_{k=1}^{\infty}\tfrac{(-1)^k}{k}\) conditionally converges.
  2. Take \((a_k)\) to be the sequence given by \(a_k=\tfrac{1}{3^k}\). The sequence is decreasing and converges to zero, so by the Alternating Series Test, \(\sum\limits_{k=1}^{\infty}\tfrac{(-1)^k}{3^k}\) is convergent. Also, \(\sum\limits_{k=1}^{\infty}\left|\tfrac{(-1)^k}{3^k}\right|=\sum\limits_{k=1}^{\infty}\left(\tfrac{1}{3}\right)^k\) is convergent because it is a geometric series with \(r=\tfrac{1}{3}\). Therefore, \(\sum_{k=1}^{\infty}\tfrac{(-1)^k}{3^k}\) is absolutely convergent.
  3. Take \((a_k)\) to be the sequence given by \(a_k=\tfrac{1}{\sqrt{3k+2}}\). The sequence is decreasing and converges to zero, so by the Alternating Series Test, \(\sum\limits_{k=1}^{\infty}\tfrac{(-1)^k}{\sqrt{3k+2}}\) converges. However, \(\sum\limits_{k=1}^\infty\left|\tfrac{(-1)^k}{\sqrt{3k+2}}\right|=\sum\limits_{k=1}^\infty\tfrac{1}{\sqrt{3k+2}}\) diverges, so the series \(\sum\limits_{k=1}^{\infty}\tfrac{(-1)^k}{\sqrt{3k+2}}\) conditionally converges.
  4. Take \((a_k)\) to be the sequence given by \(a_k=\tfrac{1}{k^2+5}\). The sequence is decreasing and converges to zero, so by the Alternating Series Test, \(\sum\limits_{k=1}^{\infty}\tfrac{(-1)^k}{k^2+5}\) is convergent. Also, \(\sum\limits_{k=1}^{\infty}\left|\tfrac{(-1)^k}{k^2+5}\right|=\sum\limits_{k=1}^{\infty}\tfrac{1}{k^2+5}\) is convergent by limit comparison test by comparing the sequence \((|a_k|)\) to the sequence \((b_k)=\left(\tfrac{1}{k^2}\right)\). So the series is absolutely convergent.

The next tests are useful as well for determining convergence and divergence.

Ratio Test

For any sequence \((a_n)\) in \(\mathbb{R}\), if there is a real number \(L\) so that \[\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=L,\] then \(\sum a_n\) is absolutely convergent if \(L\) is in \([0,1)\) and \(\sum a_n\) is divergent if \(L\) is in \((1,\infty)\) or if the limit diverges to infinity.

Practice using this test with this example. It may be helpful to rewrite \(\tfrac{a_{n+1}}{a_n}\) as \(a_{n+1}\cdot\tfrac{1}{a_n}\).

Example 15

For each of these sequences \((a_n)\), use the ratio test to determine the convergence of the series \(\sum a_n\):

  1. \(a_n=\tfrac{2^n}{n!}\)
  2. \(a_n=\tfrac{n!}{\,n^n}\)
  3. For any \(k\) in \((0,\infty)\), \(a_n=\tfrac{1}{n^k}\)

  1. Calculate the limit and get that \[\begin{align*}\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|&=\lim_{n\to\infty}\left|\frac{2^{n+1}}{(n+1)!}\cdot\frac{n!}{2^n}\right|\\&=\lim_{n\to\infty}\frac{2}{n+1}\\&=0.\end{align*}\] Because the limit is less than one, the Ratio Test states that \(\sum a_n\) is absolutely convergent.

  2. Calculate the limit and get that \[\begin{align*}\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|&=\lim_{n\to\infty}\left|\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}\right|\\&=\lim_{n\to\infty}\left|\frac{(n+1)n^n}{(n+1)^{n+1}}\right|\\&=\lim_{n\to\infty}\left|\frac{n^n}{(n+1)^n}\right|\\&=\lim_{n\to\infty}\left|\frac{n}{n+1}\right|^n\\&<1.\end{align*}\] Because the limit is less than one, the Ratio Test states that \(\sum a_n\) is absolutely convergent.

  3. Calculate the limit and get that \[\begin{align*}\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|&=\lim_{n\to\infty}\left|\frac{1}{(n+1)^{k}}\cdot n^k\right|\\&=\lim_{n\to\infty}\left|\frac{n}{n+1}\right|^k\\&=1.\end{align*}\] Because the limit equals \(1\), the test is inconclusive for all \(k\).

The other test is called the root test.

Root Test

For any sequence \((a_n)\) in \(\mathbb{R}\), if there is a real number \(L\) so that \[\lim_{n\to\infty}|a_n|^\tfrac{1}{n}=L,\] then \(\sum a_n\) is absolutely convergent if \(L\) is in \([0,1)\) and \(\sum a_n\) is divergent if \(L\) is in \((1,\infty)\) or if the limit diverges to infinity.

Practice using the root test with this example.

Example 16

For each of these series \((a_n)\), use the root test to determine the convergence of the series \(\sum a_n\):

  1. \(a_n=\tfrac{2^n}{n!}\)

  2. For any \(k\) in \((0,\infty)\), \(a_n=\tfrac{1}{\,n^k}\).

  1. The term \(|a_n|^\frac{1}{n}\) looks like this: \[|a_n|^\frac{1}{n}=\left(\frac{2^n}{n!}\right)^\frac{1}{n}=\frac{2}{\sqrt[n]{n!}}.\] To determine what the limit as \(n\) goes to infinity of this is, first notice that \(n^n\geq n!\) implies that \[n=(n^n)^\frac{1}{n}\geq (n!)^\frac{1}{n}\] and so \[(n+1)!=(n+1)\cdot n!>n\cdot n!=(n!)^\frac{1}{n}\cdot n!=(n!)^\frac{n+1}{n}.\] Take the \(\tfrac{1}{n+1}\) power to obtain that \[\left[(n+1)!\right]^\frac{1}{n+1}>(n!)^\frac{1}{n},\] which shows the sequence is increasing. Moreover, \[(n!)^\frac{1}{n}>\left(\frac{n}{2}\right)^\frac{1}{2}\quad\text{if $n\geq 4$.}\] Therefore, \[|a_n|^\frac{1}{n}=\frac{2}{\sqrt[n]{n!}}<\frac{2}{\left(\frac{n}{2}\right)^\frac{1}{2}}.\] And since the \(\frac{2}{\left(\frac{n}{2}\right)^\frac{1}{n}}\) is a null sequence, we have that \[\lim_{n\to\infty}|a_n|^\frac{1}{n}=0.\] Therefore the series absolutely converges.

  2. Note that \(\sqrt[n]{n}=1+\varepsilon_n\) with \(\varepsilon_n>0\), so \[n=(1+\varepsilon_n)^n>\frac{n(n-1)}{2}\varepsilon_n^2.\] This implies that \(0<\varepsilon_n<\sqrt{\frac{2}{n-1}}\) where the right side is a null sequence. Therefore, \((\varepsilon_n)\) is a null sequence. Thus the the limit of the terms is \[\lim_{n\to\infty}\sqrt[n]{\frac{1}{n^k}}=\lim_{n\to\infty}\left(\frac{1}{\sqrt[n]{n}}\right)^k=\lim_{n\to\infty}\left(\frac{1}{1+\varepsilon_n}\right)^k=1.\] So the test is inconclusive.

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