Why convergence forces the terms to zero
Here is the first general theorem about series, and it falls straight out of the partial-sum definition. If sum a_n converges, then a_n -> 0. The reason: the single term a_n equals s_n - s_{n-1}, the gap between consecutive partial sums. If both s_n and s_{n-1} race to the same limit s, that gap is squeezed to zero.
Theorem. If sum a_n converges then a_n -> 0.
Proof. Let s_N be the partial sums and suppose s_N -> s.
For n >= 2 we have a_n = s_n - s_(n-1).
The tail s_(n-1) also tends to s (it's the same sequence shifted).
By the algebra of limits,
a_n = s_n - s_(n-1) -> s - s = 0.
Hence a_n -> 0. QED
Contrapositive (the usable form):
If a_n does NOT tend to 0, then sum a_n diverges.
Example. sum n/(n+1): a_n = n/(n+1) -> 1, not 0.
So sum n/(n+1) diverges. (No partial-sum work needed.)In practice we use the contrapositive: this is the n-th term test for divergence. If a_n fails to approach 0, the series cannot converge, full stop. It is the cheapest check you own — always run it first.
The one-way street
The whole danger lives in that one-way street. A beginner sees a_n -> 0 and concludes the sum is finite. That is the single most common error in the subject, and the harmonic series exists precisely to refute it.
The harmonic series diverges
The series sum 1/n has terms 1/n -> 0, yet it diverges. The classic proof, due to Oresme around 1350, groups the terms into blocks and bounds each block below by 1/2. The partial sums therefore climb past every bound — slowly, but without ceiling.
Claim. sum_{n=1}^∞ 1/n diverges, even though 1/n -> 0.
Group the terms in blocks of length 1, 2, 4, 8, ...:
1
+ 1/2
+ (1/3 + 1/4) >= 1/4 + 1/4 = 1/2
+ (1/5 + 1/6 + 1/7 + 1/8) >= 4 * (1/8) = 1/2
+ (1/9 + ... + 1/16) >= 8 * (1/16) = 1/2
+ ...
Each parenthesized block of 2^(k-1) terms is >= 1/2,
because its smallest term is 1/2^k and there are 2^(k-1) of them:
2^(k-1) * (1/2^k) = 1/2.
So s_(2^k) >= 1 + k*(1/2).
The right side -> infinity as k -> infinity,
therefore the partial sums are unbounded and the series diverges. QED