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Exercise 9Calculus › Single-variable calculus › Series › Convergence testsL1Added 2 Nov 2026

Terms that do not tend to zero

#divergence test#necessary condition

Statement

Show that the following series diverge.

  1. ∑n≥1cos⁡1n\displaystyle\sum_{n\ge1}\cos\frac1n
  2. ∑n≥1(1+1n)n\displaystyle\sum_{n\ge1}\left(1+\frac1n\right)^n
  3. ∑n≥1(−1)nnn+1\displaystyle\sum_{n\ge1}(-1)^n\frac{n}{n+1}
Hint

If ∑un\sum u_n converges, then un→0u_n\to0 (because un=Sn−Sn−1u_n=S_n-S_{n-1}).

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Solution

The principle. If Sn=∑n=1kunS_n = \sum_{n=1}^k u_n converges to SS, then un=Sn−Sn−1→S−S=0u_n=S_n-S_{n-1}\to S-S=0. So if un↛0u_n\not\to0, the series diverges.

1. cos⁡1n→cos⁡0=1≠0\cos\frac1n\to\cos0=1\ne0.

2. (1+1n)n→e≠0\left(1+\frac1n\right)^n\to e\ne0.

3. ∣(−1)nnn+1∣→1\left|(-1)^n\frac n{n+1}\right|\to1, so the terms do not tend to 00.

The converse is false. un=1n→0u_n=\frac1n\to0 but ∑1n\sum\frac1n diverges. Tending to zero is *necessary* for convergence, never sufficient.

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