Summon the Heroes-3

**Xingyuan** · November 28th 2009, 04:46 PM

$Problem$ $3$

Assume $\sum_{n=1}^{\infty}\frac{a_{n}}{\sqrt{n}}$ converges

where $a_{n}>0$ and $a_{n}\geq a_{n+1}$ for $\forall$ $n \in \mathbb{N}^+$

show that: $\sum_{n=1}^{\infty}a_{n}^2$ converges

**tonio** · November 28th 2009, 06:45 PM

Originally Posted by Xingyuan

$Problem$ $3$

Assume $\sum_{n=1}^{\infty}\frac{a_{n}}{\sqrt{n}}$ converges

where $a_{n}>0$ and $a_{n}\geq a_{n+1}$ for $\forall$ $n \in \mathbb{N}^+$

show that: $\sum_{n=1}^{\infty}a_{n}^2$ converges

Excuse the doubt, but all these "summon the heroes" exercises look suspiciously similar to rather standard exercises in calculus I and II...
Are you sure these aren't "summon the guys that are going to do my homework for me" exercises? I've seen questions asked by you some few weeks ago and they did look like homework and not summonheroeds...

Tonio

**mr fantastic** · November 28th 2009, 06:56 PM

Originally Posted by Xingyuan

$Problem$ $3$

Assume $\sum_{n=1}^{\infty}\frac{a_{n}}{\sqrt{n}}$ converges

where $a_{n}>0$ and $a_{n}\geq a_{n+1}$ for $\forall$ $n \in \mathbb{N}^+$

show that: $\sum_{n=1}^{\infty}a_{n}^2$ converges

Please show your attempt at solving this question and say where you get stuck.

**Xingyuan** · November 28th 2009, 08:28 PM

I only want to make some fun in solving math problems(like playing game)

do not finish my homework.....
actually ,I am not a student now...
Thanks very much

My idea is:

As $\sum_{n=1}^{\infty}\frac{a_{n}}{\sqrt{n}}$ converges

so I think for massive(not all) $n$ , $\frac{a_{n}}{\sqrt{n}}\leq \frac{1}{n^{1+\epsilon}}$

then will give the conclusion

But now I must to prove some $a_{k}$ satisfy the condition: $\frac{a_{k}}{\sqrt{k}}\geq \frac{1}{k^{1+\epsilon}}$ , do not
make any essential contribution to the summation $\sum_{n=1}^{\infty}{a_{n}}^2$

**Laurent** · November 29th 2009, 02:11 AM

Note that $\left(\frac{a_n}{\sqrt{n}}\right)_n$ is decreasing. Therefore, the problem is equivalent to:

If $\sum_n a_n$ converges, where $(a_n)_n$ decreases, then $\sum_n n a_n^2$ converges.

(Applying this to $\frac{a_n}{\sqrt{n}}$ gives the initial statement)

Using the property from your previous "challenge", the proof is then straightforward.

(You can also directly use this property without rewriting the statement: $n\frac{a_n}{\sqrt{n}}\to_n 0$ hence $\sqrt{n} a_n\leq M$ and $a_n^2\leq a_n \frac{M}{\sqrt{n}}$ , qed)

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