$$\def\ans#1{\bbox[border:1px solid green,6pt]{#1}}$$

# Infinite Series

## Definition

If $$\{a_n\}$$ is an infinite sequence, then $\sum_{n=1}^\infty a_n = a_1 + a_2 + a_3 + a_4 + \ldots + a_n + \ldots$ is an infinite series. If an infinite series has a finite sum, it is convergent. Otherwise, it is divergent.

For instance, in the last section, we saw the example from Zeno's Paradox, where we decided that the following must be true: $\sum_{n=1}^\infty \dfrac{1}{2^n} = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \dfrac{1}{32} + \ldots = 1$ We didn't have any justification for this, other than our intuition, but in this section, we'll see it proven.

It should be clear that in order for this to be possible, the terms in the series must be getting infinitesimally small, so that adding up an infinite number of terms won't blow up to an infinite answer. Is that enough, though? We'll also see (specifically when we get to the harmonic series) that the answer is no. There are some series where the terms are getting infinitesimally small, but they aren't getting small fast enough, so the series diverges.

## Convergence: Sequence of Partial Sums

Continue with the example of the series $\sum_{n=1}^\infty \dfrac{1}{2^n}.$ How can we add up an infinite number of terms? Well, let's start with the first term: $s_1 = \dfrac{1}{2}.$ Then, add up the first two terms: $s_2 = \dfrac{1}{2}+\dfrac{1}{4} = \dfrac{3}{4}.$ What if we add up the first three terms? Or the first 4? The first 20, or 100 terms? The point is this: if, as we add up more and more terms, the sum starts to approach a limit, we can be confident that the infinite series adds up to that number. If, on the other hand, it doesn't approach a limit, we can conclude that the series diverges.

Following this process uses what's called the sequence of partial sums, $$\{s_n\}$$: \begin{align} s_1 &= \dfrac{1}{2}\\ s_2 &= \dfrac{1}{2} + \dfrac{1}{4} = \dfrac{3}{4}\\ s_3 &= \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} = \dfrac{7}{8}\\ s_4 &= \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} = \dfrac{15}{16}\\ s_5 &= \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \dfrac{1}{32} = \dfrac{31}{32}\\ &\vdots \end{align} Let's look for a pattern in $$\{s_n\}$$: $\{s_n\} = \dfrac{1}{2}, \dfrac{3}{4}, \dfrac{7}{8}, \dfrac{15}{16}, \dfrac{31}{32}, \ldots$ Notice that the numerator is always 1 less than the denominator, and the denominators are the powers of 2. Therefore, the general formula for the SOPS is $s_n = \dfrac{2^n-1}{2^n}.$

Does this sequence of partial sums approach a limit? It does, and we can show this by splitting it into two fractions: $\lim_{n \to \infty} \dfrac{2^n-1}{2^n} = \lim_{n \to \infty} \dfrac{2^n}{2^n} - \dfrac{1}{2^n} = \lim_{n \to \infty} 1-\dfrac{1}{2^n} = 1$

Therefore, by using the SOPS, we've proven that $\sum_{n=1}^\infty \dfrac{1}{2^n} = 1$ just like we expected.

### Definition

Given an infinite series $a_1 + a_2 + a_3 + a_4 + \ldots$ the sequence of partial sums is given by $$\{s_n\}$$, where $s_n = a_1 + a_2 + \ldots + a_n.$ If the sequence $$\{s_n\}$$ is convergent and $$\displaystyle\lim_{n \to \infty} s_n = s$$, then the series converges to $$s$$ as well: $\sum_{n=1}^\infty a_n = s.$ Otherwise, the series is divergent.

We use the sequence of partial sums to define what it means for a series to converge (and it's a fairly intuitive definition, since we're essentially trying to add up an infinite list of numbers by adding up more and more each time), but we don't often use it to actually identify whether or not a series converges, because the SOPS doesn't typically fit such a nice pattern. However, I'll show one more example here where it can be used.

Determine whether the following series converges. $\sum_{n=1}^\infty (-1)^n$

#### Solution

The series looks like $\sum_{n=1}^\infty (-1)^n = -1+1-1+1-1+1-\ldots$

You should verify that the sequence of partial sums, therefore, looks like \begin{align} s_1 &= -1\\ s_2 &= 0\\ s_3 &= -1\\ s_4 &= 0\\ &\vdots \end{align}

Since this sequence clearly does not converge, but simply oscillates, we can conclude that $\ans{\sum_{n=1}^\infty (-1)^n \text{ is divergent.}}$

## The Harmonic Series

The harmonic series is a specific series: $\sum_{n=1}^\infty \dfrac{1}{n} = 1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\ldots$

At first, after seeing the series at the beginning of the section, you may be tempted to conclude that this one converges as well. After all, the terms are getting smaller and smaller, infinitesimally small as the series goes on. However, it turns our that this series is divergent.

### The Harmonic Series Diverges

Here's a "proof": start by writing out the first few terms of the series.

$\sum_{n=1}^\infty \dfrac{1}{n} = 1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\ldots$

Next, group terms together as follows:

$\sum_{n=1}^\infty \dfrac{1}{n} = 1+\ans{\dfrac{1}{2}}+\ans{\dfrac{1}{3}+\dfrac{1}{4}}+\ans{\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}}+\dfrac{1}{9}+\ldots$

Replace the fractions in the boxes with smaller fractions that will make each box sum to $$1/2$$ and collapse them:

\begin{align} \sum_{n=1}^\infty \dfrac{1}{n} &= 1+\ans{\dfrac{1}{2}}+\ans{\dfrac{1}{3}+\dfrac{1}{4}}+\ans{\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}}+\dfrac{1}{9}+\ldots\\ &> 1+\ans{\dfrac{1}{2}}+\ans{\dfrac{1}{4}+\dfrac{1}{4}}+\ans{\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}}+\dfrac{1}{16}+\ldots\\ &= 1+\ans{\dfrac{1}{2}}+\ans{\dfrac{1}{2}}+\ans{\dfrac{1}{2}}+\ldots \end{align}

It should be clear that this series diverges, because there will be an infinite number of $$1/2$$s being added together. Since the harmonic series is larger than this divergent one, the harmonic series must diverge as well.

The harmonic series is a good example to remember, because there are cases where it can be used to check a result later.

## Telescoping Series

Telescoping series aren't that common, but I'll show you an example here. In general, telescoping series have the form $\sum_{n=1}^\infty \dfrac{1}{n} - \dfrac{1}{n+C}$ for some constant $$C$$. For example, consider the series $\sum_{n=1}^\infty \dfrac{1}{n} - \dfrac{1}{n+2} = 1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{5}-\dfrac{1}{7}+\ldots$

Notice what happens: the terms start to cancel each other, and the whole series collapses down (like a telescope; hence the name).

\begin{align} &1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{5}-\dfrac{1}{7}+\ldots\\ &= 1+\dfrac{1}{2}\\ &= \dfrac{3}{2} \end{align}

Therefore, this series converges:

$\sum_{n=1}^\infty \dfrac{1}{n} - \dfrac{1}{n+2} = \dfrac{3}{2}$

## Geometric Series

### Definition

A geometric series is one in which each term is a consistent multiple of the previous term. For example, the series $4+\dfrac{4}{3} + \dfrac{4}{9} + \dfrac{4}{27} + \ldots = \sum_{n=1}^\infty 4\left(\dfrac{1}{3}\right)^n$ is a geometric series, because each term is the one third of the previous term.

In general, a geometric series has the form $a+ar+ar^2+ar^3+ar^4+\ldots = \sum_{n=1}^\infty ar^{n-1} = \sum_{n=0}^\infty ar^n.$

### Convergence

The key to whether or not a geometric series converges is the value of the constant multiple $$r$$: if $$r=1$$ or $$r=-1$$, the series diverges, because it either adds the same term forever (tending toward infinity), or it oscillates between some value and zero. If $$|r| > 1$$, the series also diverges, because the terms are getting bigger and bigger, so there's no hope of convergence in that case. The only case in which a geometric series converges is if $$|r| < 1$$, in which case the series converges to $\dfrac{a}{1-r}.$ (proof)

\ans{\begin{align} \text{Summary: } &\text{If } |r| \geq 1 \text{, then the series is divergent.}\\ &\text{If } |r| < 1 \text{, then the series is convergent, and the sum is}\\ &\sum_{n=1}^\infty ar^{n-1} = \dfrac{a}{1-r} \end{align}}

### Examples

Determine whether the following series converges. $\sum_{n=1}^\infty 4\left(\dfrac{1}{3}\right)^{n-1}$

#### Solution

Here, $$a=4$$ and $$r=1/3$$. Since $$|r|<1$$, this series is convergent:

$\sum_{n=1}^\infty 4\left(\dfrac{1}{3}\right)^{n-1} = \dfrac{4}{1-\frac{1}{3}} = \ans{6}$

Determine whether the following series converges. $5-\dfrac{10}{3}+\dfrac{20}{9}-\dfrac{40}{27}+ \ldots$

#### Solution

The first term in the geometric series is $$a$$, so in this example, $$a=5$$. To get from one term to the next, multiply by $$-2/3$$, so $$r=-2/3$$. Since $$|r|<1$$, this series is convergent:

$\sum_{n=1}^\infty 5\left(-\dfrac{2}{3}\right)^{n-1} = \dfrac{5}{1+\frac{2}{3}} = \ans{3}$

Determine whether the following series converges. $\sum_{n=1}^\infty 2^{2n} \ 3^{1-n}$

#### Solution

This one is also a geometric series, although it's not written in standard form. To get it in standard form, we could manipulate it algebraically, but it's easier to simply list the first few terms, from which we can extract $$a$$ and $$r$$; remember, the first term is $$a$$ and the multiple is $$r$$.

$\sum_{n=1}^\infty 2^{2n} \ 3^{1-n} = 4+\dfrac{16}{3}+\dfrac{64}{9}+\ldots$

Therefore, $$a=4$$ and $$r=4/3$$. Since $$|r|>1$$, this series is divergent:

$\ans{\sum_{n=1}^\infty 2^{2n} \ 3^{1-n} \text{ is divergent.}}$

#### Try it yourself:

(click on a problem to show/hide its answer)