# If $\sum\limits_{n=1}^{\infty} a_n^p$ converges, for some $p>1$ then $\sum\limits_{n=1}^{\infty} \frac{a_n}{n}$ converges.

If $\{ a_n \}_{n=1}^{\infty}$ is a sequence of positive real numbers and $\sum\limits_{n=1}^{\infty} a_n^p$ converges, for some $p>1$ then $\sum\limits_{n=1}^{\infty} \frac{a_n}{n}$ converges.

Above is the full question. I have found this question in other ways, but not with a general $p$ power. I have looked at other answers that use the inequality that $|ab| \leq \frac{1}{2} (a^2 + b^2)$. I was just wondering if there was another way without this inequality?

Let $q$ be the conjugated exponent of $p$, i.e. $q=\frac{p}{p-1}$, such that $\frac{1}{p}+\frac{1}{q}=1$.
By Holder's inequality

$$\left|\sum_{n=1}^{N}\frac{a_n}{n}\right| \leq\left(\sum_{n=1}^{N}a_n^p\right)^{1/p}\left(\sum_{n=1}^{N}\frac{1}{n^q}\right)^{1/q}$$ and by letting $N\to +\infty$ we get $\left|\sum_{n\geq 1}\frac{a_n}{n}\right|\leq \zeta(q)^{1/q} \sum_{n\geq 1}a_n^p.$

• Is there a similar estimate for $0 <p\le 1$?
– user
Mar 28, 2018 at 20:09
• @user: a similar statement for $p\in(0,1)$ is known as Knopp's inequality, see page $\approx 110$ of my notes. Anyway OP's statement for $p\in(0,1)$ is trivial by $a_n\to 0$ and summation by parts. Mar 28, 2018 at 20:18
• Of course the statement itself is trivial but the statement $\sum\frac{a_n}{n}\le C(p)\sum a_n^p$ is not (if $C(p)<1$).
– user
Mar 28, 2018 at 21:01
• Ah, thank you! We did this as an application to Holder's inequality! Mar 29, 2018 at 17:39