The fastest thing you can ever check about a series
\sum_{n=1}^{\infty} a_n is whether its terms even bother to
shrink. If they do not march to zero, the series has no hope of settling on a finite total.
This is the nth-term test for divergence — a one-line sanity check that
kills off hopeless series before you waste any effort on them.
\text{If } \sum_{n=1}^{\infty} a_n \text{ converges, then } \lim_{n\to\infty} a_n = 0.
Read the other way around (the contrapositive), it becomes a test you can
actually apply: if the terms fail to die out, the series diverges.
\lim_{n\to\infty} a_n \ne 0 \;\;(\text{or the limit fails to exist}) \;\;\Longrightarrow\;\; \sum_{n=1}^{\infty} a_n \text{ diverges.}
Why convergence forces the terms to zero
The whole proof is a single clean cancellation. Write
s_n = a_1 + a_2 + \dots + a_n for the
partial sum, and
suppose the series converges to a finite limit L, meaning
s_n \to L.
Step 1 — recover a single term from neighbouring partial sums. Adding one
more term takes s_{n-1} to s_n, so the
difference is exactly the term itself:
a_n = s_n - s_{n-1}.
Step 2 — take the limit of both sides. Let
n \to \infty. The left side becomes
\lim_{n\to\infty} a_n; the right side is a limit of a difference:
\lim_{n\to\infty} a_n = \lim_{n\to\infty} \big( s_n - s_{n-1} \big).
Step 3 — split the limit of the difference. Both pieces converge, so the
limit distributes across the subtraction:
= \lim_{n\to\infty} s_n - \lim_{n\to\infty} s_{n-1}.
Step 4 — both partial-sum limits are the same number. As
n \to \infty, the index n-1 also runs
off to infinity, so s_{n-1} chases the very same limit
L:
= L - L.
Step 5 — collect. The two copies of L annihilate:
\lim_{n\to\infty} a_n = L - L = 0.
So convergence forces the terms to zero. Contrapose it and you have the test: terms
that do not vanish guarantee a divergent series.
Let \sum_{n=1}^{\infty} a_n be a series. Then:
-
Necessary condition. If
\sum a_n converges, then
\displaystyle\lim_{n\to\infty} a_n = 0.
-
The test (contrapositive). If
\displaystyle\lim_{n\to\infty} a_n \ne 0, or the limit does
not exist, then \sum a_n diverges.
-
What it cannot do. If
\displaystyle\lim_{n\to\infty} a_n = 0, the test is
inconclusive — the series may converge or diverge, and you need a sharper tool.
Heads up — the test is one-directional. Terms going to zero is
necessary for convergence, but nowhere near sufficient. The arrow only runs
one way:
\sum a_n \text{ converges} \;\Longrightarrow\; a_n \to 0, \qquad\text{but}\qquad a_n \to 0 \;\not\Longrightarrow\; \sum a_n \text{ converges.}
The most famous trap in all of analysis is the harmonic series,
\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac12 + \frac13 + \frac14 + \frac15 + \cdots.
Its terms \tfrac1n \to 0 beautifully — the divergence test is
stone silent on it. Yet the sum is +\infty. Oresme's medieval
argument groups the terms in blocks whose lengths double, and bounds each block below by
\tfrac12:
\underbrace{\tfrac13 + \tfrac14}_{>\,2\cdot\frac14 = \frac12} + \underbrace{\tfrac15 + \tfrac16 + \tfrac17 + \tfrac18}_{>\,4\cdot\frac18 = \frac12} + \cdots
Each bracket exceeds \tfrac12, and there are infinitely many
brackets, so the partial sums climb past every bound — slowly, but without limit. Terms
shrinking to zero is the price of admission to convergence, not a ticket to it.