Convergence of a Series

Convergence Is a Limit of Partial Sums

An infinite series is studied through a sequence of finite sums. For a sequence \((a_n)_{n=1}^{\infty}\), write

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

The series converges precisely when this sequence of partial sums approaches a finite real number. The individual terms matter because they build the partial sums, but the limit is a statement about the accumulated sums. This distinction gives a direct way to define convergence and a useful way to test it: examine not just one term, but every finite block of terms sufficiently far out.

Definition: The series \(\sum_{n=1}^{\infty}a_n\) converges to \(S\in\mathbb{R}\) if its partial sums satisfy \(S_N\to S\) as \(N\to\infty\). In that case, \(S\) is called the sum of the series, and we write

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

If the sequence of partial sums does not converge to a finite real number, the series diverges.

By the definition of sequence convergence, \(\sum_{n=1}^{\infty}a_n=S\) means that for every \(\varepsilon>0\), there is an integer \(N_0\) such that

N\geq N_0 \quad\Longrightarrow\quad \left|\sum_{n=1}^{N}a_n-S\right|<\varepsilon.

This says that all sufficiently long partial sums lie close to the same number \(S\). It does not say that a finite partial sum equals \(S\), nor does it say that the individual terms are themselves close to \(S\). The sum is defined by the limiting behavior of the entire sequence \((S_N)\).

The Cauchy Criterion for Series

A limit can be characterized without first knowing what it is. The Cauchy Criterion for Real Sequences says that a sequence of real numbers converges if and only if its terms eventually become arbitrarily close to one another. Applied to partial sums, the difference between two partial sums is exactly a finite block of terms, by the Block-Sum Identity from “Partial Sums.” This gives a criterion stated entirely in terms of tails of the series.

Theorem (Cauchy Criterion for Series): The series \(\sum_{n=1}^{\infty}a_n\) converges if and only if, for every \(\varepsilon>0\), there is an integer \(N_0\) such that for all integers \(p>q\geq N_0\),

\left|\sum_{n=q+1}^{p}a_n\right|<\varepsilon.

Proof. Let \(S_N=\sum_{n=1}^{N}a_n\). First suppose the series converges to \(S\). Given \(\varepsilon>0\), choose \(N_0\) so that \(N\geq N_0\) implies \(|S_N-S|<\varepsilon/2\). For \(p>q\geq N_0\), the Block-Sum Identity gives

\left|\sum_{n=q+1}^{p}a_n\right| = |S_p-S_q| \leq |S_p-S|+|S_q-S| < \frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon.

Thus every sufficiently late finite block has small absolute value.

Conversely, suppose the stated tail condition holds. Given \(\varepsilon>0\), apply it with \(\varepsilon/2\). There is an \(N_0\) such that, whenever \(p>q\geq N_0\),

|S_p-S_q| = \left|\sum_{n=q+1}^{p}a_n\right| < \frac{\varepsilon}{2}.

If \(p=q\), then \(|S_p-S_q|=0<\varepsilon/2\), so the same closeness bound holds for every \(p,q\geq N_0\), regardless of their order. Consequently \((S_N)\) is a Cauchy sequence. By the Cauchy Criterion for Real Sequences, every Cauchy sequence of real numbers converges to a finite real limit. Therefore the series converges. \(\square\)

The quantifiers are important. It is not enough that some selected blocks become small, or that one particular block has a small sum. The condition requires every block beginning after the chosen index to have small sum, no matter how many terms the block contains. This all-block requirement handles the possibility that many individually small terms might accumulate to a substantial amount.

Worked Examples

Worked Example: A Geometric Series with Ratio One Half

Consider \(\sum_{n=1}^{\infty}(1/2)^n\). The finite geometric-sum formula gives, for each \(N\geq1\),

S_N=\sum_{n=1}^{N}\left(\frac12\right)^n =1-\left(\frac12\right)^N.

For example, at \(N=1\) the formula gives \(1-\frac12=\frac12\), the first term. At \(N=2\), it gives \(1-\frac14=\frac34=\frac12+\frac14\). Since \((1/2)^N\to0\), the partial sums satisfy

\lim_{N\to\infty}S_N = \lim_{N\to\infty}\left(1-\left(\frac12\right)^N\right) = 1.

Hence the series converges to \(1\). The tail can also be calculated directly: for \(p>q\),

\sum_{n=q+1}^{p}\left(\frac12\right)^n = \left(\frac12\right)^q-\left(\frac12\right)^p,

which is positive and less than \((1/2)^q\). Given \(\varepsilon>0\), choose \(q\) large enough that \((1/2)^q<\varepsilon\). Then every such block has absolute value less than \(\varepsilon\), in agreement with the Cauchy Criterion for Series.

Worked Example: A Telescoping Series

Consider \(\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\). The partial-fraction identity is verified by

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

Therefore the \(N\)th partial sum telescopes:

S_N = \sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+1}\right) = 1-\frac{1}{N+1}.

For \(N=1\), this is \(1-\frac12=\frac12\), which equals \(1/(1\cdot2)\). For \(N=2\), it is \(1-\frac13=\frac23\), which equals \(\frac12+\frac16=\frac23\). Since \(1/(N+1)\to0\), it follows that \(S_N\to1\). Thus

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

The same cancellation verifies the tail condition explicitly. If \(p>q\), then

\sum_{n=q+1}^{p}\frac{1}{n(n+1)} = \frac{1}{q+1}-\frac{1}{p+1}.

This quantity is positive and less than \(1/(q+1)\). Choosing \(q\) so large that \(1/(q+1)<\varepsilon\) makes every finite block beyond \(q\) smaller than \(\varepsilon\).

Worked Example: A Second Telescoping Pattern

Consider \(\sum_{n=1}^{\infty}\frac{1}{n(n+2)}\). For every \(n\geq1\),

\frac12\left(\frac1n-\frac{1}{n+2}\right) = \frac12\left(\frac{n+2-n}{n(n+2)}\right) = \frac{1}{n(n+2)}.

Thus

\begin{aligned} S_N &=\frac12\sum_{n=1}^{N}\left(\frac1n-\frac{1}{n+2}\right)\\ &=\frac12\left(1+\frac12-\frac{1}{N+1}-\frac{1}{N+2}\right). \end{aligned}

At \(N=1\), this gives \(\frac12(1-\frac13)=\frac13\), equal to \(1/(1\cdot3)\). At \(N=2\), it gives \(\frac12(1+\frac12-\frac13-\frac14)=\frac{11}{24}\), and the direct sum is \(1/3+1/8=11/24\). As \(N\to\infty\), the two final fractions tend to zero, so

\sum_{n=1}^{\infty}\frac{1}{n(n+2)} = \frac12\left(1+\frac12\right) = \frac34.

This example shows why it is useful to write out the finite partial sum before taking a limit: the first terms that do not cancel determine the value of the series.

Partial Sums, Tails, and Error

When a series converges to \(S\), the difference \(S-S_N\) is called the remainder or tail after the \(N\)th partial sum. For any finite \(p>N\), the Block-Sum Identity gives

S_p-S_N=\sum_{n=N+1}^{p}a_n.

Letting \(p\) increase shows why convergence controls all sufficiently late finite tails. Conversely, the Cauchy Criterion for Series says that if all those finite tails are small, the partial sums have a limit. This is useful when the exact value of a sum is unavailable: one can sometimes establish convergence by estimating blocks, without finding the limit itself.

A basic necessary condition is the Term Test for Series Convergence, established in “What Is an Infinite Series?”: if a series converges, then \(a_n\to0\). It follows because each term is the difference \(S_n-S_{n-1}\), and consecutive terms of a convergent sequence of partial sums approach one another. This condition is not the Cauchy criterion: knowing that each individual term is small does not by itself control the sum of an arbitrarily long block. The criterion asks for control of the whole block.

Takeaway: To prove convergence, it is enough to show that every finite block sufficiently far out has arbitrarily small absolute sum. To find the sum when possible, first calculate the finite partial sums and then take their limit.

For a geometric or telescoping series, cancellation or a finite-sum formula often makes both tasks direct. In other cases, the Cauchy criterion provides the right target for estimates. The next tutorial turns to the complementary question: how to prove that a series does not converge.

Check Your Understanding

Use the definition and the Cauchy Criterion for Series to answer each question.

What sequence must converge for the series \(\sum_{n=1}^{\infty}a_n\) to converge?
Rewrite \(\sum_{n=q+1}^{p}a_n\) as a difference of partial sums.
In the Cauchy Criterion for Series, why must the condition hold for every \(p>q\geq N_0\), rather than only for one fixed-length block?
Find the \(N\)th partial sum of \(\sum_{n=1}^{\infty}(1/2)^n\), and use it to identify the sum.
For \(\sum_{n=1}^{\infty}1/[n(n+1)]\), write the finite sum from \(n=q+1\) through \(n=p\) in telescoping form.
Why does the fact that \(a_n\to0\) not state the full Cauchy Criterion for Series?

What You'll Learn