*Definition*. Let $a_1,a_2,\cdots,a_n\cdots$ be any sequence of quantities. Then the symbol

\begin{equation}

\label{eq:series}

\sum_{n=1}^\infty a_n=a_1+a_2+\cdots+a_n+\cdots

\end{equation}

is called an *infinite series*. Let

\begin{align*}

s_1&=a_1,\\

s_2&=a_1+a_2,\\

s_3&=a_1+a_2+a_3,\\

\cdots\\

s_n&=a_1+a_2+a_3+\cdots+a_n,\\

\cdots

\end{align*}

The numbers $s_n$ is called the *$n$-th partial sums* of the series \eqref{eq:series}.

*Definition*. An infinite series $\sum_{n=1}^\infty a_n$ is said to converge if the sequence of partial sums $\{s_n\}$ converges i.e. $\sum_{n=1}^\infty a_n=s<\infty$ means that for any $\epsilon>0$ there exists a positive integer $N$ such that

$$|s_n-s|<\epsilon\ \mbox{for all}\ n\geq N$$ A series which does not converge is said to diverge.

*Remark*. It should be noted that there is no unique way to define the sum of an infinite series. While the definition we use is the conventional one, there are other ways to define the sum of an infinite series. Some of the divergent series according to the conventional definition may converge with a different definition. Although it may seem outrageous it can be shown that $1+2+3+\cdots=-\frac{1}{12}$. It was first proved by the genius Indian mathematician Srinivasa Ramanujan. If you have a Netflix account, you can watch a biographical movie about Ramanujan. The movie title is *The Man Who Knew Infinity* which is based on a biography by Robert Kanigel *The Man Who Knew Infinity: A Life of the Genius Ramanujan*. I find divergent series fascinating. In case you are interested, I wrote about divergent series in blog articles here and here.

*Proposition*. If $\sum_{n=1}^\infty a_n$ converges, then $\lim_{n\to\infty}a_n=0.$

Note that the converse of the proposition is not necessarily true. See the example on harmonic series below. The proposition, more precisely its contrapositive

*If $\lim_{n\to\infty}a_n\ne 0$, then $\sum_{n=1}^\infty a_n$ diverges.*

can be used as a divergence test for series. For example, the series $\sum_{n=1}^\infty\frac{n}{n+1}$ diverges because $\lim_{n\to\infty}\frac{n}{n+1}=1\ne 0$.

*Theorem* (Cauchy’s Criterion for the Convergence of a Sequence).

A necessary and sufficient condition for the convergence of a sequence $\{a_n\}$ is that for any $\epsilon>0$ there exists a positive integer $N$ such that

$$|a_n-a_m|<\epsilon\ \mbox{for all}\ n,m\geq N.$$

*Corollary* (Cauchy’s Criterion for the Convergence of a Series).

A necessary and sufficient condition for the convergence of a series $\sum_{n=1}^\infty u_n$ is that for any $\epsilon>0$ there exists a positive integer $N$ such that

$$|s_n-s_m|<\epsilon\ \mbox{for all}\ n,m\geq N.$$

*Example* (The Geometric Series).

The geometric sequence, starting with $a$ and ratio $r$, is given by

$$a, ar,ar^2,\cdots,ar^{n-1},\cdots.$$

The $n$th partial sum is given by

$$s_n=a\frac{1-r^n}{1-r}.$$

Taking the limit as $n\to\infty$,

$$\lim_{n\to\infty}s_n=\frac{a}{1-r}\ \mbox{for}\ -1<r<1.$$

Hence the infinite geometric series converges for $-1<r<1$ and is given by

$$\sum_{n=1}^\infty ar^{n-1}=\frac{a}{1-r}.$$

On the other hand, if $r\leq -1$ or $r\geq 1$ then the infinite series diverges.

Example (The Harmonic Series).

Consider the harmonic series

$$\sum_{n=1}^\infty\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}+\cdots.$$

Group the terms as

$$1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)

+\left(\frac{1}{9}+\cdots+\frac{1}{16}\right)+\cdots.$$

Then each pair of parentheses encloses $p$ terms of the form

$$\frac{1}{p+1}+\frac{1}{p+2}+\cdots+\frac{1}{p+p}>\frac{p}{2p}=\frac{1}{2}.$$

Forming partial sums by adding the parenthetical groups one by one, we obtain

$$s_1=1, s_2=1+\frac{1}{2}, s_4>1+\frac{2}{2}, s_8>1+\frac{3}{2}, s_{16}>1+\frac{4}{2},\cdots, s_{2^n}>1+\frac{n}{2},\cdots.$$

This shows that $\lim_{n\to\infty}s_{2^n}=\infty$ and so $\{s_n\}$ diverges. Therefore, the harmonic series diverges.

*Example*. The following type of series are called *telescoping series*. #2 is left as an exercise.

- Show that

$$\sum_{n=1}^\infty\frac{1}{(2n-1)(2n+1)}=\frac{1}{2}$$*Solution*. \begin{align*}s_n&=\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)}\\&=\frac{1}{2}\sum_{k=1}^n\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)\\&=\frac{1}{2}\left(1-\frac{1}{2n+1}\right)\\&=\frac{n}{2n+1}\end{align*} and $\lim_{n\to\infty}s_n=\frac{1}{2}$. - Show that $$\sum_{n=1}^\infty\frac{1}{n(n+1)}=1$$