# Is there a positive sequence $(a_n)_{n=1}^\infty$ where both $\sum_{n=1}^\infty \frac{1}{a_n}$ and $\sum_{n=1}^\infty \frac{a_n}{n^2}$ converge?

Suppose $a_n$ is a positive sequence but not necessarily monotonic.

For the series $\sum_{n=1}^\infty \frac{1}{a_n}$ and $\sum_{n=1}^\infty \frac{a_n}{n^2}$ I can find examples where both diverge: $a_n = n$, and where one converges and the other diverges: $a_n = n^2$.

Can we find example where both converges?

• Look at $b_n=na_n$. Then sum both series, which you can do since they converge absolutely. You get $\sum \frac{1}{n}(b_n+\frac{1}{b_n})$. Now $x+\frac{1}{x}\geq 2$. – arts Nov 30 '17 at 23:27

By Cauchy-Schwarz, we have $$\sum_{n=1}^N \frac{1}{n}= \sum_{n=1}^N \frac{1}{\sqrt{a_n}} \cdot \sqrt{\frac{a_n}{n^2}} \leq \left(\sum_{n=1}^N \frac{1}{{a_n}} \right)^{1/2}\left(\sum_{n=1}^N \frac{a_n}{n^2}\right)^{1/2}$$ and the left hand side goes to $\infty$ as $N \to \infty$. So the right must as well, meaning one of the two series diverges.

No, by the Cauchy-Schartz inequality we have $$+\infty=\sum_{n=1}^\infty \frac{1}{n} = \sum_{n=1}^\infty \frac{1}{\sqrt{a_n}} \cdot \frac{\sqrt{a_n}}{n} \le \sqrt{\sum_{n=1}^\infty \frac{1}{a_n} \cdot \sum_{n=1}^\infty \frac{a_n}{n^2}}.$$

Both series cannot converge.

Suppose $\sum a_n/n^2$ converges. Since the divergent harmonic series can be written as

$$\sum_{n=1}^\infty \frac{1}{n} = \sum_{\frac{1}{n} \leqslant \frac{a_n}{n^2}} \frac{1}{n} + \sum_{\frac{1}{n} > \frac{a_n}{n^2}} \frac{1}{n},$$

and the first series on the RHS converges, it follows that the second series diverges.

Therefore,

$$\sum_{n=1}^\infty \frac{1}{a_n} > \sum_{\frac{1}{a_n} > \frac{1}{n}} \frac{1}{a_n} > \sum_{\frac{1}{a_n} > \frac{1}{n}} \frac{1}{n} = \sum_{\frac{1}{n} > \frac{a_n}{n^2}} \frac{1}{n}= +\infty$$