The Necessary Condition for Series Convergence

The Terms of a Convergent Series Must Become Small

For a series $\sum_{n=1}^{\infty}a_n$, let $S_N=\sum_{n=1}^{N}a_n$ denote its $N$th partial sum. Convergence means that the partial sums approach a finite real number. The connection between successive partial sums and individual terms is

a_n=S_n-S_{n-1},

where $S_0=0$. If the partial sums settle toward one limit, the difference between two consecutive partial sums must become small. This gives a useful restriction on every convergent series.

Theorem (Necessary Condition for Series Convergence): If the series $\sum_{n=1}^{\infty}a_n$ converges, then $a_n\to0$ as $n\to\infty$. This is the Term Test for Series Convergence, established earlier in “What Is an Infinite Series?”

The word necessary is important. The theorem says that convergence cannot occur unless the terms tend to zero. It does not say that the terms tending to zero is enough to ensure convergence. In practical use, the theorem is most decisive when its condition fails: if the terms do not tend to zero, the series must diverge.

For example, if the terms approach a nonzero number, or if their magnitudes remain bounded below by a positive constant, the series cannot converge. More generally, it is enough to find a sequence of indices tending to infinity along which the terms stay a fixed positive distance from zero. The terms need not have a limit for the necessary condition to fail.

When Failure of the Condition Proves Divergence

The contrapositive of the necessary condition gives a direct divergence test: if $a_n$ does not tend to zero, then $\sum_{n=1}^{\infty}a_n$ diverges. In terms of the definition of sequence convergence, failure to tend to zero means there is an $\varepsilon>0$ such that, however far out one goes, some later term has magnitude at least $\varepsilon$. Thus there are arbitrarily large indices $n$ for which

|a_n|\geq\varepsilon.

Each such term is the change from one partial sum to the next: $|S_n-S_{n-1}|=|a_n|$. If the partial sums converged, these consecutive differences would have to become arbitrarily small. Repeated jumps of at least $\varepsilon$ are incompatible with convergence.

Worked Example: Terms Approaching a Nonzero Limit

Consider the series with terms

a_n=\frac{2n+1}{n+1}.

Rewrite the expression as

a_n=2-\frac{1}{n+1}.

Since $1/(n+1)\to0$, the terms approach $2$, not $0$. In fact, for every $n\geq1$,

a_n=2-\frac{1}{n+1}\geq2-\frac12=\frac32.

So the terms do not approach zero; their magnitudes are at least $3/2$. By the necessary condition for series convergence, the series diverges. No calculation of its partial sums is needed.

Worked Example: Oscillating Terms That Do Not Approach Zero

Let $a_n=(-1)^n$. Then $|a_n|=1$ for every $n$, so the terms cannot tend to zero. In particular, the even-indexed terms satisfy $a_{2k}=1$ for every positive integer $k$, while the odd-indexed terms satisfy $a_{2k-1}=-1$. The terms therefore do not settle toward any limit, and certainly do not approach zero. The series $\sum_{n=1}^{\infty}(-1)^n$ diverges by the necessary condition.

The conclusion can also be seen from the partial sums: $S_{2k}=0$ and $S_{2k-1}=-1$. These two subsequences have different limits. The term test is the quicker argument here, but the calculation confirms that the terms’ persistent size corresponds to partial sums that keep making non-small changes.

Remainders Give a Quantitative View

When a series converges to $S$, define its remainder after $N$ terms by $R_N=S-S_N$, including $R_0=S$. Because $S_N\to S$, the remainders satisfy $R_N\to0$. The relationship between a term and two consecutive remainders gives more information than the necessary condition alone.

Proposition (A Term Is the Difference of Consecutive Remainders): If $\sum_{n=1}^{\infty}a_n=S$ converges and $R_N=S-S_N$, then for every $n\geq1$, $$ a_n=R_{n-1}-R_n \qquad\text{and consequently}\qquad |a_n|\leq |R_{n-1}|+|R_n|. $$

Proof. Since $S_n=S_{n-1}+a_n$, we have

R_{n-1}-R_n =(S-S_{n-1})-(S-S_n) =S_n-S_{n-1} =a_n.

Taking absolute values and applying the triangle inequality gives

|a_n|=|R_{n-1}-R_n| \leq |R_{n-1}|+|R_n|.

This inequality is useful when an estimate for the remainder is available: it translates control of the error after $N$ terms into control of the individual terms. It also describes the mechanism behind the necessary condition. A small remainder at two consecutive stages forces the term between those stages to be small.

Worked Example: A Geometric Series and Its Remainders

Consider $\sum_{n=1}^{\infty}(1/3)^n$. Its $N$th partial sum is

S_N=\frac13+\frac1{3^2}+\cdots+\frac1{3^N} =\frac12\left(1-\frac1{3^N}\right).

For instance, at $N=1$ the formula gives $\frac12(1-\frac13)=\frac13$, and at $N=2$ it gives $\frac12(1-\frac19)=\frac49=\frac13+\frac19$. Thus the series converges to $S=1/2$, and its remainder is

R_N=S-S_N=\frac{1}{2\cdot3^N}.

For every $n\geq1$, the term-remainder identity gives

R_{n-1}-R_n =\frac{1}{2\cdot3^{n-1}}-\frac{1}{2\cdot3^n} =\frac{3-1}{2\cdot3^n} =\frac1{3^n} =a_n.

The inequality in the proposition also holds, with

|R_{n-1}|+|R_n| =\frac{3}{2\cdot3^n}+\frac{1}{2\cdot3^n} =\frac{2}{3^n} \geq\frac{1}{3^n} =|a_n|.

Here the exact remainder formula shows explicitly how the tail error decreases and how each term is the change between two successive errors.

Changing Finitely Many Terms Does Not Repair the Tail

A condition about what happens as $n$ becomes large cannot be fixed by changing a finite number of initial terms. This observation is formalized by a finite-modification result. It is useful when a series has an awkward beginning but a simpler tail, or when testing whether a divergent tail can be made convergent by adjusting its first few terms.

Theorem (Finite Changes Preserve Series Convergence): Suppose two sequences $(a_n)$ and $(b_n)$ agree for all $n\geq m+1$, for some nonnegative integer $m$. Then $\sum_{n=1}^{\infty}a_n$ converges if and only if $\sum_{n=1}^{\infty}b_n$ converges. If both converge, their sums differ by the finite quantity $\sum_{n=1}^{m}(b_n-a_n)$.

Proof. Let $S_N=\sum_{n=1}^{N}a_n$ and $T_N=\sum_{n=1}^{N}b_n$. For every $N\geq m$, the terms beyond $m$ agree, so

T_N-S_N=\sum_{n=1}^{m}(b_n-a_n)=C,

where $C$ is a fixed real number. If $S_N\to S$, then $T_N=S_N+C\to S+C$, so the second series converges. Conversely, if $T_N\to T$, then $S_N=T_N-C\to T-C$, so the first series converges. When they converge, their sums differ by $C$, as claimed. The case $m=0$ is included: the sequences agree at every index, and $C=0$. $\square$

Worked Example: Altering the First Two Terms

Start with $a_n=1/[n(n+1)]$. Since

\frac{1}{n(n+1)}=\frac1n-\frac1{n+1},

the partial sums telescope:

S_N=\sum_{n=1}^{N}\left(\frac1n-\frac1{n+1}\right) =1-\frac1{N+1} =\frac{N}{N+1}.

For example, $S_1=1/2$, and $S_2=1/2+1/6=2/3$, matching the formula. As $N\to\infty$, $S_N\to1$, so the series converges.

Now define $b_1=7$, $b_2=-4$, and $b_n=1/[n(n+1)]$ for $n\geq3$. For $N\geq2$,

T_N=7-4+\sum_{n=3}^{N}\left(\frac1n-\frac1{n+1}\right) =3+\frac13-\frac1{N+1} =\frac{10}{3}-\frac1{N+1}.

For $N=2$, the sum from $n=3$ to $N$ is empty and is interpreted as zero; the displayed expression then gives $10/3-1/3=3=b_1+b_2$, as required. For $N\geq3$, the telescoping terms give $1/3-1/(N+1)$. In either case, $T_N\to10/3$. Changing the first two terms changes the sum but does not change convergence.

Why the Necessary Condition Is Not Sufficient

The Term Test is one-directional. If $a_n$ fails to approach zero, divergence follows. If $a_n\to0$, the test has no conclusion: the series might converge or diverge. To see why, consider $\sum_{n=1}^{\infty}1/\sqrt n$. Its terms tend to zero, but its partial sums are unbounded.

For each integer $j\geq1$, take the block of indices $2^{j-1}+1\leq n\leq2^j$. It contains $2^{j-1}$ terms, and every index in the block is at most $2^j$. Therefore each term satisfies $1/\sqrt n\geq1/\sqrt{2^j}$, and

\sum_{n=2^{j-1}+1}^{2^j}\frac1{\sqrt n} \geq 2^{j-1}\frac1{\sqrt{2^j}} =2^{j/2-1} \geq\frac12.

The final inequality holds because $j\geq1$ implies $2^{j/2}\geq1$. Thus every such block contributes at least $1/2$. The first $m$ blocks give $S_{2^m}\geq1+m/2$, so the partial sums are unbounded and the series diverges. Yet $1/\sqrt n\to0$. This illustrates the key distinction: individual terms becoming small is necessary, but the accumulation of infinitely many small terms still requires further analysis.

A common pitfall is to treat “the terms tend to zero” as a convergence test. It is instead a screening condition. If it fails, stop: the series diverges. If it holds, continue with an appropriate method for the series under consideration. Also, finite initial terms cannot alter whether the series converges, as the finite-change theorem shows; only the eventual behavior matters.

Check the term limit.
Determine whether $a_n\to0$. If the sequence has no limit, examine whether its magnitudes nevertheless fail to approach zero.

Use failure as a divergence proof.
If some fixed positive tolerance is exceeded by arbitrarily late terms, the necessary condition fails and the series diverges.

Do not infer convergence from success.
If $a_n\to0$, record that the necessary condition holds, but use additional reasoning to decide convergence.

Ignore finite changes when deciding convergence.
For convergence, a series and any version obtained by changing finitely many terms have the same status.

Takeaway: Convergence of $\sum_{n=1}^{\infty}a_n$ requires $a_n\to0$. Failure of this condition proves divergence immediately, but satisfying it does not prove convergence. Finite changes affect a series’ sum, not whether it converges.

Check Your Understanding

Use the necessary condition, the remainder identity, and the finite-change theorem to answer the following questions.

Why must the difference $S_n-S_{n-1}$ tend to zero if the partial sums converge?
If $|a_n|\geq1/4$ for infinitely many indices, what can you conclude about $\sum_{n=1}^{\infty}a_n$?
For a convergent series with remainder $R_N=S-S_N$, express $a_n$ using $R_{n-1}$ and $R_n$.
What does the necessary condition tell you about a series whose terms tend to zero? What does it not tell you?
If a convergent series is changed at its first three terms, does it remain convergent? How may its sum change?
In the series $\sum_{n=1}^{\infty}1/\sqrt n$, why does each block from $2^{j-1}+1$ through $2^j$ contribute at least $1/2$?