Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider a convergent sequence $\sum_{n=1}^\infty{a_n}$, with $a_n\ge 0$. Show that the series

$$\sum_{n=1}^\infty \frac {\sqrt{{a_n}}}{n}$$ also converges.

Hint: $pq\le \frac 12(p^2+q^2)$

share|cite|improve this question
up vote 9 down vote accepted

You just have to use the hint with $p:=\sqrt{a_n}$ and $q:=\frac{1}{n}$ and the knowledge of convergence of a special series.


If you have a monotonously increasing sequence $(x_n)$ which is bounded from above, then the sequence converges to $\sup_{n\in\mathbb{N}}x_n< \infty$.

Now consider the series $\sum_{n=1}^\infty \frac {\sqrt{{a_n}}}{n}$. If you say a series converges, it means that the sequence $(s_N)$ of partial sums converges, where

$$s_N=\sum_{n=1}^N \frac {\sqrt{{a_n}}}{n}.$$

Obviously, $(s_N)$ is monotonously increasing (since $s_{N+1}-s_N=\frac{\sqrt{a_{N+1}}}{N+1}>0$) and by the inequality which follows from the hint you have

$$ s_N=\sum_{n=1}^N \frac {\sqrt{{a_n}}}{n}\leq\sum_{n=1}^N\frac{1}{2}(a_n+\frac{1}{n^2})< \frac{1}{2}(\sum_{n=1}^\infty a_n+\sum_{n=1}^\infty\frac{1}{n^2})=\frac{1}{2}(A+\frac{\pi^2}{6})<\infty,$$

where $A:=\sum_{n=1}^\infty a_n$.

So $(s_N)$ is a monotonously increasing bounded sequence and hence it converges.

Or do you know the comparison test? You can argue with this here if you know it.

share|cite|improve this answer
So $$\frac{\sqrt{a_n}}{n} \le \frac12(a_n+\frac{1}{n^2})$$ Well it's obvious that $a_n$ and $\frac{1}{n^2}$ are convergent. But what theorems do I need to use to show that $$\frac{\sqrt{a_n}}{n}$$ converges? It seems so obvious, but I cannot get get the right proof because I'm confused about the details we should use. – Applied mathematician Oct 1 '12 at 21:19
You have to pay attention when you say $a_n$ converges although you mean the convergence of $\sum_{n=1}^\infty{a_n}$! I edited my answer. – Flanders Oct 2 '12 at 3:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.