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} $.
Best Answer
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.