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

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1 Answer 1

up vote 8 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.


EDIT:

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.

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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. –  MSKfdaswplwq Oct 1 '12 at 21:19
1  
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

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