Give an example of a divergent series $\sum a_n$ for which $\sum a^2_n$ also converges.

Question

So, I know an answer to this problem is $a_n = \frac{1}{n}$, but can someone explain to me WHY that's the answer?

Also, are there any alternative answers to this problem?

Answer 1

For $\alpha>1$ ($ \alpha\le1$) the series $\sum_{n=1}^{\infty} \frac{1}{n^{\alpha}}$ converges (diverges). So, take any sequence $a_n=\frac{1}{n^{\alpha}}$ with $\frac12<\alpha\le 1.$

Then, the series $$\sum _{n=1}^{\infty} a_n= \sum _{n=1}^{\infty} \frac{1}{n^{\alpha}}$$ diverges while the series $$\sum _{n=1}^{\infty} a_n^2= \sum _{n=1}^{\infty} \frac{1}{n^{2\alpha}}$$converges.

Answer 2

An (more complicated than yours) alternative is $a_n = \frac{1}{\sqrt{n+1}\log(n+1)}$. Since $\frac{1}{(n+1)\log(n+1)}$ decreases to $0$ and $\int_1^\infty \frac{dx}{(x+1)\log(x+1)}$ diverges, by the integral test, $\sum_{n = 1}^\infty \frac{1}{(n+1)\log(n+1)}$ diverges. So since $a_n > \frac{1}{(n+1)\log(n+1)}$ for all $n$, by comparison $\sum_{n = 1}^\infty a_n$ diverges. Since $a_n^2 = \frac{1}{(n+1)\log^2(n+1)}$ decreases to $0$ and $\int_1^\infty \frac{dx}{(x + 1)\log^2(x + 1)}\, dx = \frac{1}{\log 2} < \infty$, by the integral test, $\sum_{n = 1}^\infty a_n^2$ converges.

Answer 3

Take the series of primes.
$\sum\frac{1}{p}$ diverges but $\sum\frac{1}{p^2}$ converges.