What Is an Infinite Series?

From a Sequence to an Infinite Series

An infinite series begins with a sequence of real numbers, but it asks a different question from the one asked by the sequence itself. A sequence \((a_n)\) describes individual terms. A series asks whether the finite sums of those terms settle toward a finite real number as more terms are included. The central idea is that an infinite sum is defined by a limit, not by carrying out infinitely many additions at once.

We will use positive integer indices. Given a sequence \((a_n)_{n=1}^{\infty}\), its first \(N\) terms have the ordinary finite sum \(a_1+a_2+\cdots+a_N\). The sequence of these finite sums is the starting point for the definition.

Definition: Let \((a_n)_{n=1}^{\infty}\) be a sequence of real numbers. Its \(N\)th partial sum is

S_N=\sum_{n=1}^{N}a_n=a_1+a_2+\cdots+a_N.

The expression

\sum_{n=1}^{\infty}a_n

is an infinite series. The series converges to \(S\in\mathbb{R}\) if the sequence of partial sums \((S_N)\) converges to \(S\). In that case, \(S\) is called the sum of the series, and we write \(\sum_{n=1}^{\infty}a_n=S\). If the partial sums do not converge to a finite real number, the series diverges.

The notation distinguishes three related objects. The numbers \(a_n\) are the terms; the numbers \(S_N\) are the partial sums; and, if the partial sums converge, their limit is the sum of the series. The word “sum” in an infinite series therefore refers to a limit. A series can have terms that change sign, and its partial sums need not increase or decrease.

The definition also explains why a series is not merely a list of terms written with plus signs. The expression \(a_1+a_2+\cdots\) has a precise value only if its partial sums converge. The finite sum \(S_N\) is defined for each \(N\); the infinite series is assigned a sum only through the limiting behavior of the whole sequence \((S_N)\).

First Examples: Partial Sums Reveal the Meaning

Worked Example: A Geometric Series

Consider \(\sum_{n=1}^{\infty}2^{-n}\). Its \(N\)th partial sum is

S_N=\frac12+\frac14+\cdots+\frac{1}{2^N}.

To calculate it, multiply by \(1/2\) and subtract:

\begin{aligned} S_N&=\frac12+\frac14+\cdots+\frac{1}{2^N},\\ \frac12S_N&=\frac14+\frac18+\cdots+\frac{1}{2^{N+1}},\\ \frac12S_N&=\frac12-\frac{1}{2^{N+1}}. \end{aligned}

Thus \(S_N=1-2^{-N}\), as can also be checked for \(N=1\): both the original finite sum and the formula give \(1/2\). Since \(2^{-N}\) tends to zero, the partial sums tend to \(1\). By definition, the series converges and its sum is \(1\):

\sum_{n=1}^{\infty}\frac{1}{2^n}=1.

The value \(1\) is not obtained by performing infinitely many additions. It is the limit of the finite values \(1-2^{-N}\).

Worked Example: A Telescoping Series

Consider the terms \(a_n=\frac{1}{n(n+1)}\). The identity

\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}

holds because the right-hand side equals \(\frac{(n+1)-n}{n(n+1)}=\frac{1}{n(n+1)}\). Consequently, the \(N\)th partial sum is

\begin{aligned} S_N &=\sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+1}\right)\\ &=\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+ \left(\frac1N-\frac{1}{N+1}\right)\\ &=1-\frac{1}{N+1}. \end{aligned}

Every intermediate term cancels with its neighbor; the displayed expression shows that the only uncancelled terms are \(1\) and \(-1/(N+1)\). Since \(1/(N+1)\) tends to zero, \(S_N\) tends to \(1\). Therefore,

\sum_{n=1}^{\infty}\frac{1}{n(n+1)}=1.

The cancellation is finite at every stage. The conclusion about the infinite series follows only after taking the limit of the resulting partial-sum formula.

Worked Example: Alternating Partial Sums

Consider \(1-1+1-1+\cdots\), represented by \(a_n=(-1)^{n+1}\). The first partial sums are \(S_1=1\), \(S_2=0\), \(S_3=1\), and \(S_4=0\). More generally, \(S_N=1\) when \(N\) is odd and \(S_N=0\) when \(N\) is even. The partial sums do not approach one finite limit: the odd-indexed partial sums remain \(1\), while the even-indexed partial sums remain \(0\). Since these two values differ, the sequence \((S_N)\) does not converge. Hence this series diverges.

This example shows why the terms alone cannot be added informally to assign a sum. Grouping the expression as \((1-1)+(1-1)+\cdots\) might suggest \(0\), while grouping it as \(1+(-1+1)+(-1+1)+\cdots\) might suggest \(1\). The definition avoids this ambiguity: it fixes the order of the terms and tests the limit of the corresponding partial sums.

A Necessary Condition for Convergence

The definition gives an immediate relation between a term and two neighboring partial sums: for \(N\geq2\), \(a_N=S_N-S_{N-1}\). If the partial sums settle toward one limit, their successive differences must approach zero. This yields an important first test for divergence.

Theorem (Term Test for Series Convergence): If the series \(\sum_{n=1}^{\infty}a_n\) converges, then \(a_n\) tends to zero as \(n\) tends to infinity.

Proof. Suppose the series converges to \(S\). By definition, \(S_N\to S\) as \(N\to\infty\). For \(N\geq2\), the definition of the partial sums gives

a_N=S_N-S_{N-1}.

As \(N\to\infty\), \(S_N\to S\), and \(S_{N-1}\to S\) as well, since shifting the index of a convergent sequence does not change its limit. The limit laws for sequences therefore give

\lim_{N\to\infty}a_N = \lim_{N\to\infty}(S_N-S_{N-1}) = S-S = 0.

Thus the terms of every convergent series tend to zero. \(\square\)

The theorem supplies a one-way test. If the terms fail to tend to zero, the series must diverge. But the theorem does not say that terms tending to zero guarantee convergence; the next example makes that limitation explicit.

Worked Example: Terms Tend to Zero but the Series Diverges

Consider the harmonic series \(\sum_{n=1}^{\infty}1/n\). Its terms tend to zero. Nevertheless, its partial sums are unbounded. To see this, group terms between successive powers of two. For each integer \(k\geq0\), the block from \(n=2^k+1\) to \(n=2^{k+1}\) contains \(2^k\) terms. Each term in the block is at least \(1/2^{k+1}\), so the block contributes at least

2^k\cdot\frac{1}{2^{k+1}}=\frac12.

For example, the block for \(k=0\) consists of \(1/2\), and the block for \(k=1\) consists of \(1/3+1/4\), which is at least \(1/2\). Including the first term \(1\), this estimate gives, for every integer \(m\geq1\),

S_{2^m} = 1+\sum_{k=0}^{m-1}\sum_{n=2^k+1}^{2^{k+1}}\frac1n \geq 1+\sum_{k=0}^{m-1}\frac12 = 1+\frac{m}{2}.

The equality in the block decomposition accounts for all terms from \(2\) through \(2^m\), with no overlap or omission. As \(m\) increases, \(1+m/2\) is unbounded, so the subsequence \(S_{2^m}\) is unbounded. A convergent sequence is bounded; consequently, the partial sums cannot converge to a finite real number. The harmonic series diverges, despite \(1/n\to0\).

Linearity for Convergent Series

Finite sums obey familiar algebraic rules. The definition lets us transfer those rules to infinite series when the relevant series converge: take the finite partial sums first, apply the finite algebra, and then pass to their limits.

Theorem (Linearity of Convergent Series): Suppose \(\sum_{n=1}^{\infty}a_n=A\) and \(\sum_{n=1}^{\infty}b_n=B\) converge, and let \(\alpha,\beta\in\mathbb{R}\). Then \(\sum_{n=1}^{\infty}(\alpha a_n+\beta b_n)\) converges and

\sum_{n=1}^{\infty}(\alpha a_n+\beta b_n)=\alpha A+\beta B.

Proof. Let \(S_N=\sum_{n=1}^{N}a_n\) and \(T_N=\sum_{n=1}^{N}b_n\). The \(N\)th partial sum of the combined series is a finite sum, so distributivity gives

\sum_{n=1}^{N}(\alpha a_n+\beta b_n) = \alpha\sum_{n=1}^{N}a_n+\beta\sum_{n=1}^{N}b_n = \alpha S_N+\beta T_N.

By hypothesis, \(S_N\to A\) and \(T_N\to B\). The limit laws for sequences imply \(\alpha S_N+\beta T_N\to\alpha A+\beta B\). The partial sums of the combined series therefore converge to \(\alpha A+\beta B\), which proves both convergence and the stated formula. This reasoning includes zero values of either coefficient, since the same finite identity and limit law still apply. \(\square\)

Worked Example: Combining Two Convergent Series

The geometric-series calculation above gives \(\sum_{n=1}^{\infty}2^{-n}=1\), and the telescoping calculation gives \(\sum_{n=1}^{\infty}1/(n(n+1))=1\). Both series converge, so linearity applies to their term-by-term sum:

\sum_{n=1}^{\infty} \left(\frac{1}{2^n}+\frac{1}{n(n+1)}\right) = 1+1=2.

For a check at the level of partial sums, the \(N\)th partial sum of the combined series is

\left(1-\frac{1}{2^N}\right)+\left(1-\frac{1}{N+1}\right) = 2-\frac{1}{2^N}-\frac{1}{N+1}.

Both subtracted terms tend to zero, so these partial sums tend to \(2\), in agreement with the theorem. The calculation illustrates the order of the reasoning: first establish convergence of the two original series, then apply linearity.

What the Definition Does—and Does Not—Promise

The partial-sum definition gives a stable foundation for later tests and calculations. It also prevents several common errors:

Do not confuse terms and partial sums. Convergence of a series means convergence of \((S_N)\), not merely convergence of \((a_n)\).
Use the term test only in the valid direction. If \(a_n\) does not tend to zero, the series diverges. If \(a_n\to0\), the series may still converge or diverge.
Keep the order of the terms fixed. The partial sums use the first \(N\) terms in their stated order. Changing how terms are grouped or rearranged is not part of the definition given here.
Apply algebra only when convergence is known. The linearity theorem assumes both component series converge; it does not authorize combining divergent series as if their sums were ordinary real numbers.

The geometric and telescoping examples converged because their partial sums had explicit formulas with finite limits. The alternating example diverged because its partial sums did not settle to one value. The harmonic example showed that terms tending to zero is not enough. Together, these cases emphasize the central question for every series: what happens to its partial sums as \(N\) grows?

The next step is to study partial sums more closely. Their exact formulas, differences, and limiting behavior provide the main tools for deciding whether particular infinite series converge and for determining their sums when they do.

Check Your Understanding

Use the definition through partial sums to answer each question.

What sequence must converge for the series \(\sum_{n=1}^{\infty}a_n\) to have a sum?
Why does convergence of a series imply that its terms tend to zero?
What does the term test allow you to conclude if the terms fail to tend to zero?
Why does \(1/n\to0\) not establish convergence of the harmonic series?
If two series converge to \(A\) and \(B\), what is the sum of the series with terms \(\alpha a_n+\beta b_n\)?

What You'll Learn