Prove a Series Convergence Theorem

From Sequence Convergence to Series Convergence

A series is not a new kind of limit: it is the limit of a sequence of partial sums. That connection makes the Cauchy Criterion for Real Sequences an effective tool for proving a series converges. In this tutorial, we use it to prove a central result: if the series of absolute values converges, then the original series converges.

The key idea is to control every finite tail of the original series by the corresponding tail of the series of absolute values. Since those absolute-value tails become arbitrarily small, the partial sums of the original series satisfy the Cauchy condition. Completeness of the real numbers then supplies their limit.

Definition: Let $(a_n)$ be a sequence of real numbers, and define its $N$th partial sum by $$ S_N=\sum_{n=1}^{N}a_n. $$ The series $\sum_{n=1}^{\infty}a_n$ converges to $S\in\mathbb R$ if $S_N\to S$ as $N\to\infty$. It converges absolutely if the series $\sum_{n=1}^{\infty}|a_n|$ converges.

A series can converge even when its terms have both signs; its partial sums may approach a finite limit through cancellation. Absolute convergence is a stronger condition: it requires convergence even after that cancellation is removed. We first express convergence directly in terms of finite tails.

The Cauchy Criterion for Series

Theorem (Cauchy Criterion for Series): A series $\sum_{n=1}^{\infty}a_n$ converges if and only if, for every $\varepsilon>0$, there is a positive integer $N$ such that $$ \left|\sum_{k=n+1}^{m}a_k\right|<\varepsilon $$ whenever $m>n\geq N$.

Proof. Let $S_N=\sum_{k=1}^{N}a_k$. For integers $m>n$, subtracting the two partial sums cancels their common first $n$ terms:

S_m-S_n =\sum_{k=1}^{m}a_k-\sum_{k=1}^{n}a_k =\sum_{k=n+1}^{m}a_k.

Therefore, the stated finite-tail condition is exactly the Cauchy condition for the sequence $(S_N)$. If the series converges, then its partial sums converge, so they are Cauchy by the Convergence Implies the Cauchy Condition theorem. The displayed identity gives the required tail condition.

Conversely, suppose the finite-tail condition holds. Given $\varepsilon>0$, choose $N$ as in that condition. If $p,q\geq N$ and $p>q$, then $\lvert S_p-S_q\rvert=\lvert\sum_{k=q+1}^{p}a_k\rvert<\varepsilon$. If $q>p$, the same bound follows by reversing the difference, and if $p=q$, the difference is zero. Thus $(S_N)$ is a Cauchy sequence. By the Cauchy Criterion for Real Sequences, it converges in $\mathbb R$, which means that the series converges. $\square$

The order of the indices matters: the criterion controls every tail beginning after a sufficiently large $n$, not just tails beginning at one selected index. This uniform control over all $m>n$ is what verifies the Cauchy condition.

Absolute Convergence Implies Convergence

For any finite set of real numbers, the triangle inequality bounds the absolute value of their sum by the sum of their absolute values. The following proof applies that inequality to every finite tail, then uses the Cauchy Criterion for Series.

Theorem (Absolute Convergence Implies Convergence): If $\sum_{n=1}^{\infty}|a_n|$ converges, then $\sum_{n=1}^{\infty}a_n$ converges.

Proof. Let $T_N=\sum_{k=1}^{N}|a_k|$. By hypothesis, $(T_N)$ converges, so it is Cauchy. Let $\varepsilon>0$. There is a positive integer $N$ such that $\lvert T_m-T_n\rvert<\varepsilon$ whenever $m>n\geq N$. For these indices, the finite triangle inequality gives

\left|\sum_{k=n+1}^{m}a_k\right| \leq \sum_{k=n+1}^{m}|a_k| =T_m-T_n <\varepsilon.

The Cauchy Criterion for Series now shows that $\sum_{n=1}^{\infty}a_n$ converges. $\square$

The proof needs no formula for the sum and no special pattern in the signs of the terms. The only estimate is applied to finite sums, where the triangle inequality is available directly. The Cauchy criterion then turns that estimate into convergence.

Worked Example: An Alternating Geometric Series

Consider $\sum_{n=1}^{\infty}(-1)^{n-1}/4^n$. Its absolute-value series is geometric, and its $N$th partial sum is

\sum_{n=1}^{N}\frac{1}{4^n} =\frac{\frac14(1-(\frac14)^N)}{1-\frac14} =\frac13\left(1-\frac{1}{4^N}\right).

As $N\to\infty$, $4^{-N}\to0$, so these partial sums converge to $1/3$. The absolute-value series therefore converges. By the theorem, the original series converges as well. In fact, its partial sums satisfy

\sum_{n=1}^{N}\frac{(-1)^{n-1}}{4^n} =\frac{\frac14(1-(-\frac14)^N)}{1+\frac14} =\frac15\left(1-(-\frac14)^N\right),

which tend to $1/5$. This example verifies the sufficient condition by first evaluating the absolute-value series.

Worked Example: A Signed Telescoping Series

Let $a_n=(-1)^n/[n(n+1)]$ for $n\geq1$. For every $n$,

|a_n|=\frac{1}{n(n+1)} =\frac1n-\frac{1}{n+1},

because $\frac1n-\frac{1}{n+1}=\frac{(n+1)-n}{n(n+1)}=\frac{1}{n(n+1)}$. Consequently, the $N$th partial sum of the absolute-value series is

\sum_{n=1}^{N}|a_n| =\sum_{n=1}^{N}\left(\frac1n-\frac{1}{n+1}\right) =1-\frac{1}{N+1}.

The partial sums tend to $1$, so the series of absolute values converges. The theorem proves that $\sum_{n=1}^{\infty}(-1)^n/[n(n+1)]$ converges. The calculation establishes absolute convergence without requiring an evaluation of the signed series.

A Tail Estimate for the Error

The same finite-sum inequality gives more than a convergence proof. It bounds the distance between the sum of an absolutely convergent series and any of its partial sums.

Theorem (Absolute-Convergence Tail Estimate): Suppose $\sum_{n=1}^{\infty}a_n$ converges absolutely to $S$. Then, for every positive integer $N$, $$ |S-S_N|\leq \sum_{k=N+1}^{\infty}|a_k|. $$

Proof. For every $m>N$, the finite triangle inequality gives

\left|\sum_{k=N+1}^{m}a_k\right| \leq \sum_{k=N+1}^{m}|a_k|.

As $m\to\infty$, the sum on the left inside the absolute value tends to $S-S_N$: the first $N$ terms of the partial sum have been subtracted. The right side tends to $\sum_{k=N+1}^{\infty}|a_k|$, since it is a tail of a convergent series. By the Absolute-Value Limit Theorem, the absolute values on the left tend to $|S-S_N|$. Passing to the limit in the inequality, using that order is preserved under limits, proves the estimate. $\square$

Worked Example: Choosing a Partial Sum to Meet an Error Bound

For the alternating geometric series $\sum_{n=1}^{\infty}(-1)^{n-1}/4^n$, the absolute tail after the $N$th term is

\sum_{k=N+1}^{\infty}\frac{1}{4^k} =\frac{4^{-(N+1)}}{1-\frac14} =\frac{1}{3\cdot4^N}.

The tail estimate gives $\lvert S-S_N\rvert\leq 1/(3\cdot4^N)$. To make this error at most $1/100$, it is enough to choose $N$ so that $1/(3\cdot4^N)\leq1/100$, or $4^N\geq100/3$. Since $4^3=64\geq100/3$, taking $N=3$ suffices. The partial sum is

S_3=\frac14-\frac1{16}+\frac1{64} =\frac{16-4+1}{64} =\frac{13}{64}.

Thus $\lvert S-13/64\rvert\leq1/192<1/100$. The bound certifies the accuracy of this approximation without needing the exact value of the sum.

Why Absolute Convergence Is Sufficient, Not Necessary

It is important not to reverse the theorem: convergence does not imply absolute convergence. Cancellation can make the partial sums settle down even when the sum of the absolute values grows without bound. The alternating harmonic series provides a precise example.

Worked Example: The Alternating Harmonic Series

Consider $\sum_{n=1}^{\infty}(-1)^{n+1}/n$, and write $S_N$ for its partial sums. The even partial sums can be grouped into positive pairs:

S_{2m}=\sum_{j=1}^{m}\left(\frac{1}{2j-1}-\frac{1}{2j}\right).

Each pair is positive, since $\frac{1}{2j-1}-\frac{1}{2j}=\frac{1}{(2j-1)2j}>0$. Thus $(S_{2m})$ is increasing: adding the next pair increases its value. It is also bounded above by $1$, because

S_{2m} =1-\sum_{j=1}^{m-1}\left(\frac{1}{2j}-\frac{1}{2j+1}\right)-\frac{1}{2m} \leq1.

By the Monotone Convergence Theorem, the even partial sums converge to some $L$. The odd partial sums satisfy $S_{2m+1}=S_{2m}+1/(2m+1)$. Since $1/(2m+1)\to0$, the Stability of Sequence Limits Under Vanishing Perturbations theorem shows that $S_{2m+1}\to L$ as well. The even and odd partial sums together form the full sequence of partial sums, so $S_N\to L$. The alternating harmonic series converges.

It does not converge absolutely. The partial sums of its absolute-value series are harmonic sums. For each positive integer $r$, group the terms from $2^{j-1}+1$ through $2^j$, for $1\leq j\leq r$. Each such block has $2^{j-1}$ terms, and each term is at least $1/2^j$. Therefore,

\sum_{n=1}^{2^r}\frac1n \geq 1+\sum_{j=1}^{r}2^{j-1}\frac{1}{2^j} =1+\frac r2.

These partial sums are unbounded as $r$ increases, so the harmonic series diverges. The alternating harmonic series thus converges without converging absolutely.

This distinction is useful in practice. Absolute convergence lets us control tails without analyzing how positive and negative terms cancel. When absolute convergence is unavailable, convergence may still hold, but it requires another argument that uses the structure of the terms. The theorem proved here supplies a robust sufficient condition, not a characterization of every convergent series.

Form the partial sums.
To prove a series converges, regard it as the sequence $S_N=\sum_{n=1}^{N}a_n$.

Control finite tails.
Use the triangle inequality to bound $\lvert\sum_{k=n+1}^{m}a_k\rvert$ by $\sum_{k=n+1}^{m}|a_k|$.

Apply the series Cauchy criterion.
If the tails of the absolute-value series become arbitrarily small, the original partial sums are Cauchy and hence converge.

Check Your Understanding

Use the definitions and proofs above to answer these questions.

How does the difference $S_m-S_n$ relate to a finite tail of the series?
Where does the triangle inequality enter the proof that absolute convergence implies convergence?
What additional information does the Absolute-Convergence Tail Estimate provide beyond convergence alone?
Why does convergence of the alternating harmonic series not contradict the theorem?
For a series that converges absolutely, what quantity can be used to choose a partial sum with a prescribed error bound?