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

Question

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} $.

Answer 1

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.

Answer 2

Hint: Use the Cauchy-Schwarz inequality.

Answer 3

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 $\epsilon \leq 1$

Answer 4

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).