# How do people sense that Cauchy-Schwarz inequality will be used in this question?

I was reading this question on this website.

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 $\frac{1}{2}$.

I spent a lot of time thinking about the problem but after lot of mental struggle I gave up. I looked at the answers posted and it was a simple application of Cauchy-Schwarz inequality.

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

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

In short my question is what are some red flags to notice which signals that Cauchy-Schwarz inequality can be used to solve the problem.

I know the theorem and proof as well.

P.S. Sorry for my English. Feel free to edit.

You've got an arbitrary convergence series $\sum a_n$, and you're trying to figure out the convergence of $\sum \frac{\sqrt{a_n}}{n^p}$. Two things strike me about that series.
Firstly, the terms are related to $\sqrt{a_n}$, rather than just $a_n$. That will change convergence, in particular, it will make convergence slower (or make convergence not happen at all). If we want to relate this series back to $\sum a_n$, it'd be really handy if there was some inequality that naturally involved looking at the sum of squared terms, such as the Cauchy-Schwarz inequality.