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Let $\{ a_n\} $ be a sequence of non-negative real numbers such that the series $ \sum_{n=1}^{\infty}a_n $ is convergent.

If $p $ is a real number such that the series $ \sum{\frac{\sqrt a_n}{n^p}} $ diverges, then

(A) $p$ must be strictly less than $\frac{1}{2} $

(B) $p$ must be strictly less than or equal to $\frac{1}{2} $

(C) $p$ must be strictly less than or equal to 1 but can be greater than $\frac{1}{2} $

(D) $p$ must be strictly less than 1 but can be greater than or equal to $\frac{1}{2} $.

So how to approach? The numerator converges is given. And now from $p$ series we know for $p \le 1$ $\sum{\frac{1}{1^p}} $ diverges. So how to bring the range closer to $\frac{1}{2} $.

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Being a bit more explicit, the Cauchy-Schwarz inequality and the assumptions imply

$$\infty = \sum_{n=1}^\infty \frac{a_n^{1/2}}{n^p} \leq \left ( \sum_{n=1}^\infty a_n \right )^{1/2} \left ( \sum_{n=1}^\infty \frac{1}{n^{2p}} \right )^{1/2}.$$

Since you know the second sum converges, the third sum must diverge. Finish from there.

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  • $\begingroup$ Solving the third sum is coming back to $ \sum{\frac{1}{n^p}}$ which diverges only when $p \le 1$. I apologise if I'm doing some mistake here. How to reduce it more closer to $\frac 1 2$ @Ian $\endgroup$
    – N S
    Mar 6 '15 at 23:27
  • $\begingroup$ @NS The third sum diverges if and only if $2p \leq 1$ meaning $p \leq 1/2$. $\endgroup$
    – Ian
    Mar 6 '15 at 23:29
  • $\begingroup$ why arent you considering the 1/2 whole power? Yes now I understood. Nice manipulation. But wont 1/2 and 2 cancel? $\endgroup$
    – N S
    Mar 6 '15 at 23:31
  • $\begingroup$ @NS What? The third sum has a power of $2p $. It diverges if and only if that power is less than or equal to 1. This occurs if and only if the original $ p $ is less than or equal to 1/2. $\endgroup$
    – Ian
    Mar 6 '15 at 23:34
  • $\begingroup$ Yes that's obvious. I get that. I was saying the whole third sum is raised to power of 1/2 and n is raised to power of 2p. So that's what I was saying. But now I get my mistake. The whole summation is raised to power of 1/2. not individuals. I apologise for asking such silly things $\endgroup$
    – N S
    Mar 6 '15 at 23:37
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Hint: Use the Cauchy-Schwarz inequality.

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  • $\begingroup$ I thought of that. But I'm not able to conclude. Actually I'm looking for a more intuitive easy to undestand solution. @Potato $\endgroup$
    – N S
    Mar 6 '15 at 23:06
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A good idea is to consider the serie

$$ \sum \frac{1}{n\ln^{1+\epsilon}(x)} $$

Then, if

$$ \sum \frac{1}{n^p} \sqrt{\frac{1}{n\ln^{1+\epsilon}(x)}} $$ diverge, it means that

$$ \sum \frac{1}{n^{p+\frac{1}{2}}\ln^{\frac{1+\epsilon}{2}}(x) } $$

diverge, the serie diverge if $p=\frac{1}{2}$ and $\epsilon \leq 1$

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  • $\begingroup$ This one looks really tough to understand. Anything more intuitive? $\endgroup$
    – N S
    Mar 6 '15 at 23:08
  • $\begingroup$ what is the $ \epsilon $ for? $\endgroup$
    – N S
    Mar 6 '15 at 23:10
  • $\begingroup$ Actually, I made a mistake, this prove that $p= \frac{1}{2}$ the serie diverge if $\epsilon \leq 1$ $\endgroup$
    – Tryss
    Mar 6 '15 at 23:12
  • $\begingroup$ @NS : any positive number. It's a classical convergent serie and an interesting case between $\frac{1}{n}$ and $\frac{1}{n^{1+\epsilon}}$ $\endgroup$
    – Tryss
    Mar 6 '15 at 23:16
  • $\begingroup$ I don't see how that's relevant to the question $\endgroup$
    – Ant
    Mar 6 '15 at 23:26
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Can we apply the root test ? The nth term and (n+1)th term of the series is $|T_{n}|=\frac{\sqrt{a_{n}}}{(n)^{p}}$ and $|T_{n+1}|=\frac{\sqrt{a_{n+1}}}{(n+1)^{p}}$ respectively. Then by root test $|\frac{T_{n+1}}{T_n}|= \sqrt{\frac{a_{n+1}}{a_{n}}} \frac{n^p}{(n+1)^p}$ , where the series $\sum_{i=1}^\infty a_{n}$ is convergent , so $\frac{a_{n+1}}{a_{n}} < 1$, and $\frac{n^p}{(n+1)^p}$ is $> 1 $ if $p \leq 1$. (Series $\sum_{i=1}^\infty \frac{1}{n}$ is divergent).

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  • $\begingroup$ @Ian will root test work here? Why did'nt you treat the expression as $|x.y|\leq |x||y|$ instead of using $|x.y|^{\frac{1}{2}}\leq |x|^{\frac{1}{2}}|y|^{\frac{1}{2}}$ ? $\endgroup$ Apr 28 '16 at 2:30

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