Infinite Series: Adding Without an End

By Vivek Sharma·MATHEMATICS·Ch. 1 · Infinite Series and Power SeriesSunday, October 18, 2026· 14 min read

Imagine walking halfway to a wall, then half the remaining distance, then half again. You keep adding smaller distances. Can infinitely many pieces fit into a finite distance? Infinite series give us a precise way to answer that question. You only need fractions and basic algebra to begin; we will explain limits as we go.

1. A sequence lists; a series adds

A sequence is an ordered list of numbers. We write $a_n$ for the term in position $n$ : the subscript labels the position; it is not a power. For example, $a_n=1/2^n$ for $n=1,2,3,\ldots$ gives $1/2,1/4,1/8,\ldots$ .

A series adds the terms of a sequence. The capital Greek letter $\Sigma$ (sigma) means ‘add’. The expression below says: start at $n=1$ , substitute each whole-number value of $n$ , and keep adding.

\sum_{n=1}^{\infty}a_n=a_1+a_2+a_3+\cdots

\sum_{n=1}^{\infty}\frac{1}{2^n}=\frac12+\frac14+\frac18+\cdots

Remember: A sequence is a list of terms; a series is their sum.

2. Add finitely many terms first

We cannot finish infinitely many additions one at a time. Instead, add the first $N$ terms and call the result the partial sum $S_N$ . Then ask what happens as $N$ grows.

A limit is the value these partial sums approach: we can make the distance from that value as small as we want by taking enough terms. If the partial sums approach a finite number $S$ , the series converges, and its sum is $S$ . Otherwise, it diverges.

S_N=\sum_{n=1}^{N}a_n

\sum_{n=1}^{\infty}a_n=S\quad\text{means}\quad\lim_{N\to\infty}S_N=S

Worked example · Halving the distance

For the series $1/2+1/4+1/8+\cdots$ , the first four partial sums are $S_1=1/2$ , $S_2=3/4$ , $S_3=7/8$ , and $S_4=15/16$ .
After $N$ terms, $S_N=1-2^{-N}$ . The gap to $1$ is $2^{-N}$ , which approaches zero. The series therefore converges to $1$ .
After ten terms, the gap is $1/1024$ , less than $0.001$ . Partial sums can give useful approximations before we take a limit.

Remember: Convergence describes the limit of the partial sums, not just the individual terms.

3. The geometric series: a reusable pattern

A geometric series multiplies each term by the same common ratio $r$ . If the first term is $a$ , its terms are $a, ar, ar^2, \ldots$ . Starting the index at zero makes $ar^0=a$ the first term.

For $r\ne1$ , multiply a partial sum by $r$ and subtract. All the middle terms cancel, giving the finite-sum formula below. If $|r|<1$ , then $r^N$ approaches zero, giving the infinite-sum formula. Here $|r|$ means the distance of $r$ from zero.

S_N=a+ar+\cdots+ar^{N-1}=\frac{a(1-r^N)}{1-r}

\sum_{n=0}^{\infty}ar^n=\frac{a}{1-r}\qquad (|r|<1)

Worked example · A shrinking contribution

To sum $3+1+1/3+1/9+\cdots$ , identify $a=3$ and $r=1/3$ .
Since $|1/3|<1$ , the sum exists. Substitute into the formula: $3/(1-1/3)=9/2=4.5$ .
A negative ratio can also work. For $1-1/2+1/4-\cdots$ , the ratio is $-1/2$ , so the sum is $1/(1+1/2)=2/3$ . The partial sums alternate around this limit.

Remember: For a nonzero first term, a geometric series converges exactly when the common ratio has magnitude less than one.

4. Smaller terms are necessary, but not enough

If a series converges, its terms must approach zero. So if the terms do not approach zero, the series diverges. This is called the term test for divergence.

The reverse is false. The harmonic series has terms $1/n$ that approach zero, but the sum grows without bound. Grouping its terms explains why: each group below contributes at least $1/2$ , and there are infinitely many groups.

Divergence can also mean oscillation. The partial sums of $1-1+1-1+\cdots$ alternate between $1$ and $0$ , so they have no single limit.

\sum_{n=1}^{\infty}\frac1n=1+\frac12+\underbrace{\left(\frac13+\frac14\right)}_{\geq1/2}+\underbrace{\left(\frac15+\frac16+\frac17+\frac18\right)}_{\geq1/2}+\cdots

Remember: Terms failing to approach zero prove divergence; terms approaching zero leave the question open.

5. A small toolkit for deciding convergence

Start with the term test. Then look for a geometric pattern. If neither settles the question, choose a test suited to the terms.

For positive-term series, the p-series rule says that $\sum_{n=1}^{\infty}1/n^p$ converges when $p>1$ and diverges when $p\leq1$ . Thus $\sum1/n^2$ converges, while the harmonic series ( $p=1$ ) diverges. This rule decides convergence; it does not automatically give the sum.

Comparison test: if $0\leq a_n\leq b_n$ for all sufficiently large $n$ and $\sum b_n$ converges, then $\sum a_n$ converges too. For example, $1/(n^2+1)\leq1/n^2$ , so its series converges.

Ratio test: compute the limit $L$ below, when it exists and the ratios are defined eventually. If $L<1$ , the series converges absolutely (even the sum of the absolute values converges). If $L>1$ , including an infinite limit, the series diverges. If $L=1$ , the test gives no answer.

For alternating signs, $\sum_{n=1}^{\infty}(-1)^{n-1}b_n$ converges if $b_n>0$ , the $b_n$ decrease eventually, and $b_n\to0$ . For example, $1-1/2+1/3-1/4+\cdots$ converges, although the sum of its absolute values is the divergent harmonic series. This is conditional convergence.

L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|

Worked example · When the ratio test cannot decide

For $a_n=1/n$ , the ratio $a_{n+1}/a_n=n/(n+1)$ tends to $1$ . The harmonic series diverges.
For $a_n=1/n^2$ , the ratio is $n^2/(n+1)^2$ , which also tends to $1$ . This series converges by the p-series rule.
The same ratio-test result can occur for both outcomes. When $L=1$ , switch tests.

Remember: Choose a test from the pattern of the terms. Proving convergence and finding an exact sum are separate tasks.