Series and basic convergence

Add the terms of a sequence one after another, forever, and you get an infinite series. Sometimes that endless sum settles down to a finite number — the series converges — and sometimes it grows without bound or oscillates forever — it diverges. The whole subject of infinite series, which occupies much of this unit, is built on this single fork in the road, and everything else (comparison tests, ratio tests, power series) is machinery for deciding which branch a given series falls on. This article lays the foundation: how a series is defined through its partial sums, the geometric series as the one family you can sum exactly by hand, the harmonic series as the classic cautionary tale, and the necessary (but never sufficient) $n$ -th term test that every other test builds on.

The core insight

An infinite series converges exactly when its sequence of partial sums approaches a finite limit, and a term sequence that fails to shrink to zero guarantees divergence — but shrinking to zero alone guarantees nothing.

The core idea

Partial sums define the series. Given a sequence $\{a_n\}_{n=1}^\infty$ , the infinite series $\sum_{n=1}^{\infty} a_n$ is not itself a sum you can compute term by term forever — it is defined as the limit of the sequence of partial sums,

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

where: $a_n$ is the $n$ -th term of the underlying sequence and $S_N$ is the sum of the first $N$ terms only, a perfectly ordinary finite sum. The series converges to $S$ if $\lim_{N\to\infty} S_N = S$ exists as a finite number, and diverges otherwise. This reframing — from "adding forever" to "take the limit of a sequence of finite sums" — is what turns an intuitively suspicious idea into something you can actually analyze with the limit tools from earlier in the course.

The geometric series is the one family summed exactly. For $\sum_{n=0}^{\infty} ar^n$ with $a \ne 0$ ,

\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \quad \text{if } |r|<1, \qquad \text{diverges if } |r|\ge 1.

where: $a$ is the first term (at $n=0$ ) and $r$ is the common ratio between consecutive terms. Because its partial sum has the closed form $S_N = a\dfrac{1-r^N}{1-r}$ , you can watch $r^N \to 0$ directly when $|r|<1$ and see the limit $a/(1-r)$ emerge — this is the one series family in the whole unit where convergence and the exact sum are both fully transparent.

The $n$ -th term test only ever proves divergence. If $\sum a_n$ converges, then necessarily $a_n \to 0$ . Equivalently, if the terms do not shrink to zero (or their limit does not exist), the series must diverge. Crucially, the converse is false: $a_n \to 0$ is necessary but not sufficient, so this test can rule a series out but can never rule one in.

The harmonic series is the standard proof that "terms shrink" is not enough. $\sum_{n=1}^{\infty} \dfrac{1}{n}$ has $a_n \to 0$ , yet the series diverges — its partial sums grow like $\ln N$ , climbing without bound, just agonizingly slowly (it takes over $12{,}000$ terms to exceed $S_N = 10$ ). Keep this example next to the $n$ -th term test at all times; it is precisely the case the test cannot handle, which is why comparison and ratio tests exist at all.

Figure

Two bar charts of partial sums, one for a geometric series levelling off at a finite height and one for the harmonic series climbing without bound as more bars are added

Derivation ANecessary condition for convergence

The goal is to establish, with nothing beyond the algebra of limits already available from the sequences unit, that a convergent series' terms must vanish — and to make the logical direction of that statement airtight, since misusing it (treating it as if it also proved convergence) is the single most common error students make with series.

Step 1. Assume $\sum_{n=1}^{\infty} a_n$ converges to a finite sum $S$ . By definition this means the sequence of partial sums satisfies $\lim_{N\to\infty} S_N = S$ .

Step 2. Write the $n$ -th term as a difference of consecutive partial sums. For $n \ge 2$ , $a_n = S_n - S_{n-1}$ , since $S_n = a_1 + \cdots + a_n$ and $S_{n-1} = a_1 + \cdots + a_{n-1}$ differ by exactly the last term added.

Step 3. Take the limit of both sides as $n \to \infty$ . Since $\{S_n\}$ and $\{S_{n-1}\}$ are both just shifted copies of the same convergent sequence, both tend to $S$ : $\lim_{n\to\infty} S_n = S$ and $\lim_{n\to\infty} S_{n-1} = S$ . By the difference law for limits,

\lim_{n\to\infty} a_n = \lim_{n\to\infty} (S_n - S_{n-1}) = S - S = 0.

Step 4. Read off the logical structure. Convergence of the series forces $a_n \to 0$ ; this is the necessary condition. Taking the contrapositive gives the test's practical form: if $a_n \not\to 0$ , the series cannot converge. But the proof never shows the reverse implication — nothing here says $a_n \to 0$ forces convergence — which is exactly why the harmonic series (terms $\to 0$ , series diverges) is not a contradiction of this result but a demonstration of its one-directional limit.

Worked example

Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{3}{4^n}$ converges, and if so, find its sum; then contrast it with $\displaystyle\sum_{n=1}^{\infty} \frac{n}{2n+1}$ .

Step 1. Recognize the first series' form. Write $\dfrac{3}{4^n} = 3\left(\dfrac{1}{4}\right)^n$ , a geometric series with first term (at $n=1$ ) $a_1 = 3/4$ and common ratio $r = 1/4$ .

Step 2. Check convergence and sum it. Since $|r| = 1/4 < 1$ , the series converges, and using $\sum_{n=1}^{\infty} a_1 r^{n-1} = \dfrac{a_1}{1-r}$ ,

\sum_{n=1}^{\infty} \frac{3}{4^n} = \frac{3/4}{1 - 1/4} = \frac{3/4}{3/4} = 1.

Step 3. Now the second series: try the $n$ -th term test first, since it is the cheapest check available.

\lim_{n\to\infty} \frac{n}{2n+1} = \lim_{n\to\infty} \frac{1}{2 + 1/n} = \frac{1}{2}.

Step 4. Conclude. The terms tend to $1/2 \ne 0$ , so by the contrapositive of the necessary condition, $\sum \dfrac{n}{2n+1}$ diverges — this took one line, whereas the geometric series required the full closed-form sum. Final answers: $\sum 3/4^n = 1$ (converges); $\sum n/(2n+1)$ diverges.

Competitive stretch

A telescoping series collapses its own partial sum, giving a third route to a clean answer. Take $\sum_{n=1}^{\infty}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)$ : writing out $S_N$ , every interior term cancels against its neighbour, leaving $S_N = 1 - \dfrac{1}{N+1}$ . As $N \to \infty$ , $S_N \to 1$ , so the series converges to $1$ — recognizing the telescoping pattern is often faster than any general test, when the terms happen to factor that way.

Common traps

Believing $a_n \to 0$ guarantees convergence. It does not — the harmonic series is the standard counterexample: terms vanish, the series still diverges.
Misidentifying the ratio in a geometric series, especially with signs, e.g. $\sum (-2/3)^n$ has $r = -2/3$ ; students sometimes drop the sign or misplace the starting index, which changes the first term $a$ used in $a/(1-r)$ .
Applying the $n$ -th term test backwards, i.e. concluding "converges" once $a_n \to 0$ . The test is one-directional: it only ever proves divergence, never convergence.
Confusing the sequence $\{a_n\}$ with the sequence of partial sums $\{S_N\}$ . "The sequence converges" and "the series converges" are different statements about different sequences; always be explicit about which one is meant.

Ask the tutor: Want practice spotting whether a given series is geometric, telescoping, or a job for the $n$ -th term test — or working through why the harmonic series diverges despite its shrinking terms?

Why bother defining a series through partial sums instead of just adding infinitely many numbers directly?

Because "adding infinitely many numbers" is not a well-defined arithmetic operation on its own — addition is only guaranteed to work for finitely many terms. Partial sums sidestep the problem entirely: $S_N$ is an ordinary finite sum for every $N$ , so $\{S_N\}$ is an ordinary sequence, and you already know from the sequences unit what it means for a sequence to have a limit. Defining the series' value as that limit converts an intuitively suspicious infinite process into a completely rigorous limit statement, with no new machinery required.

If a series fails the n-th term test, do I even need to try other tests?

No — failing the $n$ -th term test (terms not tending to zero) is a complete proof of divergence on its own, so you can stop immediately and write "diverges by the $n$ -th term test." The trap runs the other way: if the terms do tend to zero, that tells you nothing yet, and you must still reach for comparison, ratio, root, or integral tests (covered next) to settle the question. Always run the $n$ -th term test first simply because it is nearly free, but never treat passing it as an answer.

Does removing or changing the first hundred terms of a series change whether it converges?

No. Convergence or divergence is entirely a statement about the tail of the sequence — how $a_n$ behaves as $n \to \infty$ — so changing, adding, or deleting any finite number of terms never flips convergence to divergence or vice versa. It can, however, change the value of the sum if the series does converge, since you have literally altered some of the numbers being added. This is why proofs often say "for $n$ sufficiently large" and freely ignore small- $n$ behaviour when only convergence, not the exact sum, is at stake.

Series and basic convergence

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads