## Section6.2Series

While much more can be said about sequences, we now turn to our principal interest, series. Recall that a series, roughly speaking, is the sum of a sequence: If $\ds\{a_n\}_{n=0}^\infty$ is a sequence then we can construct a series by adding up all the terms in the sequence:

\begin{equation*} \sum_{n=0}^\infty a_n=a_0+a_1+a_2+\cdots \end{equation*}
###### Definition6.23. Infinite Series.

Given an infinite sequence $\{a_n\}_{n=0}^{\infty}\text{,}$ the sum

\begin{equation*} \sum_{n=0}^{\infty} a_n = a_0 + a_1 + a_2 + \dots \end{equation*}

is called an infinite series.

Associated with a series is a second sequence, called the sequence of partial sums $\ds\{s_n\}_{n=0}^\infty\text{,}$ where the $n$-th partial sum $s_n$ always terminates with the $n$-th term $a_n$ of the sequence:

\begin{equation*} s_0=a_0, s_1=a_0+a_1, s_2=a_0+a_1+a_2,\dots \end{equation*}
###### Definition6.24. Sequence of Partial Sums.

Given an infinite series $\ds\sum_{n=0}^{\infty} a_n\text{,}$ the sequence of partial sums is $\{s_n\}_{n=0}^{\infty}\text{,}$ where the $n$-th partial sum is defined as

\begin{equation*} s_n = \sum_{i=0}^n a_i = a_0+a_1+a_2 + \dots + a_n\text{.} \end{equation*}

Again, we are interested in the behaviour of the series, and whether its terms sum to a finite number or not. We therefore define convergence and divergence for a series.

###### Definition6.25. Convergence and Divergence of a Series.

Given an infinite series $\ds\sum_{n=0}^{\infty} a_n\text{,}$ the series converges if the associated sequence of partial sums converges; otherwise, the series diverges .

###### Definition6.26. Geometric Series.

If $\{kx^n\}_{n=0}^{\infty}$ is a geometric sequence, then the associated series $\sum_{i=0}^{\infty}kx^i$ is called a geometric series.

Note: An infinite series $\ds\sum_{n=0}^{\infty}a_n$ can also be denoted by $\ds\sum_n a_n$ or $\ds\sum a_n\text{,}$ where it is understood that $n$ is an integer that runs over all admissible values of the index. However, if we are interested in the actual sum of the series, then it is pertinent to identify the lower bound of the series.

A typical partial sum is

\begin{equation*} s_n=k+kx+kx^2+kx^3+\cdots+kx^n=k(1+x+x^2+x^3+\cdots+x^n)\text{.} \end{equation*}

We note that for $x \neq 1\text{,}$

\begin{align*} s_n(1-x)\amp =k(1+x+x^2+x^3+\cdots+x^n)(1-x)\\ \amp =k(1+x+x^2+x^3+\cdots+x^n)1-k(1+x+x^2+x^3+\cdots+x^{n-1}+x^n)x\\ \amp =k(1+x+x^2+x^3+\cdots+x^n-x-x^2-x^3-\cdots-x^n-x^{n+1})\\ \amp =k(1-x^{n+1}) \end{align*}

so

\begin{align*} s_n(1-x)\amp =k(1-x^{n+1})\\ s_n\amp =k{1-x^{n+1}\over 1-x}\text{.} \end{align*}
1. If $|x|\lt 1\text{,}$ $\ds\lim_{n\to\infty}x^n=0$ so

\begin{equation*} \lim_{n\to\infty}s_n=\lim_{n\to\infty}k{1-x^{n+1}\over 1-x}= k{1\over 1-x}\text{.} \end{equation*}

Thus, when $|x|\lt 1$ the geometric series converges to $k/(1-x)\text{.}$

1. If $|x| > 1\text{,}$ then

\begin{equation*} \lim_{n\to\infty} x^n = \infty \end{equation*}

and so the series diverges.

2. If $|x| = 1\text{,}$ then

\begin{equation*} |s_n| = \sum_{i=0}^{n} 1 = n+1\text{.} \end{equation*}

Therefore,

\begin{equation*} \lim_{n\to\infty} |s_n| = \lim_{n\to\infty} (n+1) = \infty \end{equation*}

and so the series diverges.

###### Example6.28. Summing a Geometric Series.

Given the series $\ds\sum_{n=0}^{\infty} \frac{1}{2^n}\text{,}$ determine the following:

1. The partial sum $s_n\text{.}$

2. The sum of the series.

Solution

We recognize that the series is geometric with $k=1$ and $x=1/2\text{.}$

1. $\ds{s_n={1-(1/2)^{n+1}\over 1-1/2}={2^{n+1}-1\over 2^n}=2-{1\over 2^n}}$

2. $\ds{\sum_{n=0}^\infty {1\over 2^n} = {1\over 1-1/2} = 2.}$

We began the chapter with the series

\begin{equation*} \sum_{n=1}^\infty {1\over 2^n}\text{,} \end{equation*}

namely, the geometric series without the first term $1\text{.}$ Each partial sum of this series is 1 less than the corresponding partial sum for the geometric series, so of course the limit is also one less than the value of the geometric series, that is,

\begin{equation*} \sum_{n=1}^\infty {1\over 2^n}=1\text{.} \end{equation*}

It is not hard to see that the following theorem follows from Theorem 6.6.

Note:

1. When $c$ is non-zero, the converse of the first part of this theorem is also true. That is, if $\sum ca_n$ is convergent, then $\sum a_n$ is also convergent; if $\sum ca_n$ converges then $\frac{1}{c}\sum ca_n$ must converge.

2. On the other hand, the converse of the second part of the theorem is not true. For example, if $a_n=1$ and $b_n=-1\text{,}$ then $\sum a_n+\sum b_n=\sum 0=0$ converges, but each of $\sum a_n$ and $\sum b_n$ diverges.

In general, the sequence of partial sums $\ds s_n$ is harder to understand and analyze than the sequence of terms $\ds a_n\text{,}$ and it is difficult to determine whether series converge and if so to what. The following result will let us deal with some simple cases easily.

Since $\sum a_n$ converges, $\ds\lim_{n\to\infty}s_n=L$ and $\ds\lim_{n\to\infty}s_{n-1}=L\text{,}$ because this really says the same thing but “renumbers” the terms. By Theorem 6.6,

\begin{equation*} \lim_{n\to\infty} (s_{n}-s_{n-1})= \lim_{n\to\infty} s_{n}-\lim_{n\to\infty}s_{n-1}=L-L=0\text{.} \end{equation*}

But

\begin{equation*} s_{n}-s_{n-1}=(a_0+a_1+a_2+\cdots+a_n)-(a_0+a_1+a_2+\cdots+a_{n-1}) =a_n\text{,} \end{equation*}

so as desired $\ds\lim_{n\to\infty}a_n=0\text{.}$

Note:

1. This theorem presents an easy Divergence Test when we use the contrapositive form: Given a series $\sum a_n\text{,}$ if the limit $\ds\lim_{n\to\infty}a_n$ does not exist or has a value other than zero, the series diverges. This result is captured in the next theorem called the $n$-th Term Test. Often, this theorem is referred to as the Divergence Test.

2. Note well that the converse is not true: If $\ds\lim_{n\to\infty}a_n=0$ then the series does not necessarily converge. One example is the so-called harmonic series, as will be shown in Example 6.33.

Consider the statement of the theorem in contrapositive form:

\begin{equation*} \text{ If } \ \ds\sum_{n=1}^{\infty}a_n\text{ converges, then } \lim_{n\to\infty}a_n=0\text{.} \end{equation*}

If $s_n$ are the partial sums of the series, then the assumption that the series converges gives us

\begin{equation*} \ds\lim_{n\to\infty}s_n=s \end{equation*}

for some number $s\text{.}$ Then

\begin{equation*} \ds\lim_{n\to\infty}a_n=\lim_{n\to\infty}(s_n-s_{n-1})=\lim_{n\to\infty}s_n-\lim_{n\to\infty}s_{n-1}=s-s=0\text{.} \end{equation*}
###### Example6.32. Demonstrating Divergence.

Show that $\ds\sum_{n=1}^\infty {n\over n+1}$ diverges.

Solution

We compute the limit:

\begin{equation*} \lim _{n\to\infty}{n\over n+1}=1\not=0\text{.} \end{equation*}

Looking at the first few terms perhaps makes it clear that the series has no chance of converging:

\begin{equation*} {1\over2}+{2\over3}+{3\over4}+{4\over5}+\cdots \end{equation*}

will just get larger and larger; indeed, after a bit longer the series starts to look very much like $\cdots+1+1+1+1+\cdots\text{,}$ and of course if we add up enough 1's we can make the sum as large as we desire.

###### Example6.33. Harmonic Series.

Show that $\ds\sum_{n=1}^\infty {1\over n}$ diverges.

Solution

Here the theorem does not apply: $\ds\lim _{n\to\infty} 1/n=0\text{,}$ so it looks like perhaps the series converges. Indeed, if you have the fortitude (or the software) to add up the first 1000 terms you will find that

\begin{equation*} \sum_{n=1}^{1000} {1\over n}\approx 7.49\text{,} \end{equation*}

so it might be reasonable to speculate that the series converges to something in the neighborhood of 10. But in fact the partial sums do go to infinity; they just get big very, very slowly. Consider the following:

\begin{equation*} \begin{aligned} \amp\ds 1+{1\over 2}+{1\over 3}+{1\over 4} > 1+{1\over 2}+{1\over 4}+{1\over 4} = 1+{1\over 2}+{1\over 2}\\ \amp \ds 1+{1\over 2}+{1\over 3}+{1\over 4}+ {1\over 5}+{1\over 6}+{1\over 7}+{1\over 8} > 1+{1\over 2}+{1\over 4}+{1\over 4}+{1\over 8}+{1\over 8}+{1\over 8}+{1\over 8} = 1+{1\over 2}+{1\over 2}+{1\over 2}\\ \amp \ds 1+{1\over 2}+{1\over 3}+\cdots+{1\over16}> 1+{1\over 2}+{1\over 4}+{1\over 4}+{1\over 8}+\cdots+{1\over 8}+{1\over16}+\cdots +{1\over16} =1+{1\over 2}+{1\over 2}+{1\over 2}+{1\over 2}\end{aligned} \end{equation*}

and so on. By swallowing up more and more terms we can always manage to add at least another $1/2$ to the sum, and by adding enough of these we can make the partial sums as big as we like. In fact, it's not hard to see from this pattern that

\begin{equation*} 1+{1\over 2}+{1\over 3}+\cdots+{1\over 2^n} > 1+{n\over 2}\text{,} \end{equation*}

so to make sure the sum is over 100, for example, we'd add up terms until we get to around $\ds 1/2^{198}\text{,}$ that is, about $\ds 4\cdot 10^{59}$ terms.

###### Definition6.34. Harmonic Series.

A series of the form

\begin{equation*} \sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} + \dots \end{equation*}

is called a harmonic series.

Note: We will often make use of the fact that the first few (e.g. any finite number of) terms in a series are irrelevant when determining whether it will converge. In other words, $\ds\sum_{n=0}^{\infty}a_n$ converges if and only if $\ds\sum_{n=N}^{\infty}a_n$ converges for some $N\geq 1\text{.}$

##### Exercises for Section 6.2.

Explain why the following series diverge.

1. $\ds\sum_{n=1}^\infty {n^2\over 2n^2+1}$

$\ds\lim_{n\to\infty} n^2/(2n^2+1)=1/2$
Solution

$\lim\limits_{n\to\infty} \dfrac{n^2}{2n^2+1} = \lim\limits_{n\to\infty} \dfrac{1}{2+\frac{1}{n^2}} = \dfrac{1}{2} \neq 0\text{,}$ so the series $\displaystyle\sum_{n=0}^{\infty} \frac{n^2}{2n^2+1}$ diverges.

2. $\ds\sum_{n=1}^\infty {5\over 2^{1/n}+14}$

$\ds\lim_{n\to\infty} 5/(2^{1/n}+14)=1/3$
Solution

$\lim\limits_{n\to\infty} \dfrac{5}{2^{1/n}+14} = \dfrac{5}{15} = \dfrac{1}{3} \neq 0\text{,}$ so the series $\displaystyle\sum_{n=0}^{\infty}\frac{5}{2^{1/n}+14}$ diverges.

3. $\ds\sum_{n=1}^\infty {3\over n}$

$\ds\sum_{n=1}^\infty {1\over n}$ diverges, so $\ds\sum_{n=1}^\infty 3{1\over n}$ diverges
Solution

We notice that $\displaystyle\sum_{n=1}^{\infty} \dfrac{3}{n} = 3 \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n}\text{,}$ which diverges since the harmonic series diverges.

Compute the sum of the following series, if it converges.

1. $\ds\sum_{n=0}^\infty \left({4\over (-3)^n}- {3\over 3^n}\right)$

$-3/2$
Solution
We first rewrite
\begin{equation*} \sum_{n=0}^{\infty} \left(\frac{4}{(-3)^n} - \frac{3}{3^n}\right) = \sum_{n=0}^{\infty} \frac{4}{(-3)^n} - \sum_{n=0}^{\infty} \frac{3}{3^n}. \end{equation*}
We now notice that these are both geometric series with $|x| = 1/3 \le 1$ , and so the series converges to:
\begin{equation*} \sum_{n=0}^{\infty} \frac{4}{(-3)^n} - \sum_{n=0}^{\infty} \frac{3}{3^n} = \frac{4}{1-(-1/3)} - \frac{3}{1-(1/3)} = \frac{-3}{2}. \end{equation*}
2. $\ds\sum_{n=0}^\infty \left({3\over 2^n}+ {4\over 5^n}\right)$

$11$
Solution
We first rewrite
\begin{equation*} \sum_{n=0}^{\infty} \left(\frac{3}{2^n} + \frac{4}{5^n}\right) = \sum_{n=0}^{\infty} \frac{3}{2^n} +\sum_{n=0}^{\infty} \frac{4}{5^n}. \end{equation*}
We notice that these are both geometrix series with $|x| = 1/2$ and $|x| = 1/5\text{,}$ respectively. So the series converges to:
\begin{equation*} \sum_{n=0}^{\infty} \left(\frac{3}{2^n} + \frac{4}{5^n}\right) = \frac{3}{1-1/2} + \frac{4}{4-1/5} = 11. \end{equation*}
3. $\ds\sum_{n=0}^\infty {4^{n+1}\over 5^n}$

$20$
Solution

The series

\begin{equation*} \sum_{n=0}^{\infty} \frac{4^{n+1}}{5^n} = \sum_{n=0}^{\infty} 4\left(\frac{4}{5}\right)^n \end{equation*}

is a geometric series. Since $\frac{4}{5} \lt 1\text{,}$ the series converges with

\begin{equation*} \sum_{n=0}^{\infty} \frac{4^{n+1}}{5^n} = \frac{4}{1-\frac{4}{5}} = 20\text{.} \end{equation*}
4. $\ds\sum_{n=0}^\infty {3^{n+1}\over 7^{n+1}}$

$3/4$
Solution

We first rewrite the series as

\begin{equation*} \sum_{n=0}^{\infty} \frac{3^{n+1}}{7^{n+1}} = \sum_{n=0}^{\infty} \frac{3}{7} \left(\frac{3}{7}\right)^n\text{.} \end{equation*}

This is a geometric series, and since $\frac{3}{7} \lt 1\text{,}$ the series converges:

\begin{equation*} \sum_{n=0}^{\infty} \frac{3^{n+1}}{7^{n+1}} = \frac{\frac{3}{7}}{1-\frac{3}{7}} = \frac{3}{4}\text{.} \end{equation*}
5. $\ds\sum_{n=1}^\infty \left({3\over 5}\right)^n$

$3/2$
Solution

We notice that this is a convergent geometric series, since $3/5 \lt 1\text{.}$ Now let

\begin{equation*} \sum_{n=1}^{\infty} \left(\frac{3}{5}\right)^n = \sum_{n=0}^{\infty} \left(\frac{3}{5}\right)^n - \left(\frac{3}{5}\right)^0 = \sum_{n=0}^{\infty} \left(\frac{3}{5}\right)^n -1\text{.} \end{equation*}

Using our result for the convergence of geometric series, we find

\begin{equation*} \sum_{n=1}^{\infty} \left(\frac{3}{5}\right)^n = \frac{1}{1-\frac{3}{5}} - 1 = \frac{3}{2}\text{.} \end{equation*}
6. $\ds\sum_{n=1}^\infty {3^n\over 5^{n+1}}$

$1/2$