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

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 $\frac{1}{2}$.

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.

• 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
– N S
Mar 6 '15 at 23:27
• @NS The third sum diverges if and only if $2p \leq 1$ meaning $p \leq 1/2$.
– Ian
Mar 6 '15 at 23:29
• why arent you considering the 1/2 whole power? Yes now I understood. Nice manipulation. But wont 1/2 and 2 cancel?
– N S
Mar 6 '15 at 23:31
• @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.
– Ian
Mar 6 '15 at 23:34
• 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
– N S
Mar 6 '15 at 23:37

Hint: Use the Cauchy-Schwarz inequality.

• I thought of that. But I'm not able to conclude. Actually I'm looking for a more intuitive easy to undestand solution. @Potato
– N S
Mar 6 '15 at 23:06

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 $\epsilon \leq 1$

• This one looks really tough to understand. Anything more intuitive?
– N S
Mar 6 '15 at 23:08
• what is the $\epsilon$ for?
– N S
Mar 6 '15 at 23:10
• Actually, I made a mistake, this prove that $p= \frac{1}{2}$ the serie diverge if $\epsilon \leq 1$ Mar 6 '15 at 23:12
• @NS : any positive number. It's a classical convergent serie and an interesting case between $\frac{1}{n}$ and $\frac{1}{n^{1+\epsilon}}$ Mar 6 '15 at 23:16
• I don't see how that's relevant to the question
– Ant
Mar 6 '15 at 23:26

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).

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