Series

An infinite sequence $a_n$ can be used to form an infinite series, or simply series, by summing its entries:

$$\sum _{n=1}^{\mathrm{\infty }}{a}_n=a_1+a_2+a_3+\cdots$$

We can define a sequence $s_n$ of partial sums of the infinite series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ by assigning

$$s_n=\sum _{k=1}^n{a}_n=a_1+a_2+\cdots +a_n.$$

The series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ is said to be convergent and to have a sum $m$ if the sequence of partial sums $s_n$ converges to $m.$ If the sequence $s_n$ diverges, then the series is said to diverge.

Absolute Convergence

The series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ is said to converge absolutely if $\sum _{n=1}^{\mathrm{\infty }}\Vert a_n\Vert$ converges. The infinite series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ is said to converge conditionally if it converges but $\sum _{n=1}^{\mathrm{\infty }}\Vert a_n\Vert$ diverges.

Linearity of Summation of Convergent Series

Theorem: If $c$ is a real number and the series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ and $\sum _{n=1}^{\mathrm{\infty }}{b}_n$ are convergent, then

$$\sum _{n=1}^{\mathrm{\infty }}(a_n+c\cdot b_n)=\sum _{n=1}^{\mathrm{\infty }}{a}_n+c\sum _{n=1}^{\mathrm{\infty }}{b}_n.$$

Convergence Tests

Before considering the sum of a series, we need to know if it converges or not. There are many tests for convergence; we will start with a test for divergence.

Divergence Test

Theorem: The divergence test says that if an infinite series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges, then $\underset{n\to \mathrm{\infty }}{lim}{a}_n=0.$ Therefore, if $\underset{n\to \mathrm{\infty }}{lim}{a}_n\ne 0$ (if the limit diverges or converges to a value other than 0), the series diverges.

Proof: Suppose $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges with sum $m$ and let $s_n=\sum _{k=1}^n{a}_n.$ Then, it follows that

$$\underset{n\to \mathrm{\infty }}{lim}{a}_n=\underset{n\to \mathrm{\infty }}{lim}(s_n-s_{n-1})=\underset{n\to \mathrm{\infty }}{lim}{s}_n-\underset{n\to \mathrm{\infty }}{lim}{s}_{n-1}=m-m=0.◻$$

Note that $a_n$ converging to $0$ is a necessary condition for $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ to converge, but it is not sufficient.

Geometric Series Test

Theorem: The geometric series test says that the geometric series $\sum _{n=1}^{\mathrm{\infty }}{r}^n$ converges if and diverges otherwise. If then $\sum _{n=1}^{\mathrm{\infty }}{r}^n=\frac{r}{1-r}.$

Proof: Suppose Note that

$$s_n=(r+r^2+\cdots +r^n)=r(1+r+r^2+\cdots +r^{n-1})=r\frac{1-r^n}{1-r}.$$

(That last step is a rabbit out of a hat but can be shown using long division.) Now, if we take the limit as n goes to infinity we have

$$\underset{n\to \mathrm{\infty }}{lim}{s}_n=\underset{n\to \mathrm{\infty }}{lim}r\frac{1-r^n}{1-r}=r\frac{1}{1-r}=\frac{r}{1-r}.◻$$

Note that when $\Vert r\Vert \ge 1,$ the series diverges.

Bounded Series with Nonnegative Terms

Theorem: Suppose $a_n\ge 0$ for all $n\in \mathbb{N}.$ Then the series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges iff the sequence of partial sums $s_n=\sum _{k=1}^n{a}_k$ is bounded.

Proof: Assume the sequence of partial sums $s_n$ is bounded. Since all terms of $a_n$ are nonnegative, $s_n$ is increasing, and by the Monotone Convergence Theorem, must converge, meaning $\sum _{n=1}^{\mathrm{\infty }}{a}_n=\underset{n\to \mathrm{\infty }}{lim}{s}_n$ converges. On the other hand, assume $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges. Then $s_n$ is a convergent sequence and is therefore bounded (see sequences).

$2^n$ Test

Theorem: Suppose $a_n\ge a_{n+1}\ge 0$ for all $n\in \mathbb{N}$ (i.e. $a_n$ is nonnegative and decreasing.) Then the series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges iff the series $\sum _{n=1}^{\mathrm{\infty }}{2}^n{a}_{2n}$ converges.

$p$-Series Test

Theorem: The $p$-series $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^p}$ converges if and diverges otherwise.

Alternating Series Test

Theorem: Suppose $a_n\ge a_{n+1}$ for all natural numbers $n$ and that $\underset{n\to \mathrm{\infty }}{lim}{a}_n=0.$ Then the alternating series $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}{a}_n$ converges.

Absolute Value Test

Theorem: If the series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges absolutely then it simply converges too.

Comparison Test

Theorem: Suppose $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ and $\sum _{n=1}^{\mathrm{\infty }}{b}_n$ are series for which $\Vert a_n\Vert \le \Vert b_n\Vert$ for all natural $n.$ Then if $\sum _{n=1}^{\mathrm{\infty }}{b}_n$ converges absolutely, so does $\sum _{n=1}^{\mathrm{\infty }}{a}_n.$ If $\sum _{n=1}^{\mathrm{\infty }}\Vert a_n\Vert$ diverges, so does $\sum _{n=1}^{\mathrm{\infty }}\Vert b_n\Vert .$

Limit Ratio Test

Theorem: For the series $\sum _{n=1}^{\mathrm{\infty }}{a}_n,$ let

$$L=\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|.$$

Then, if $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges absolutely. If $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ diverges.

Root Test

Theorem: For the series $\sum _{n=1}^{\mathrm{\infty }}{a}_n,$ let $L=\underset{n\to \mathrm{\infty }}{lim sup}\Vert a_n{\Vert }^{1/n}.$ Then if $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges absolutely. If $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ diverges.