Let $a_n$ and $b_n$ be two sequences of real numbers such that series $a_n^2$ and $b_n^2$ converge. Then the series $a_nb_n$ [duplicate]

Let $a_n$ and $b_n$ be two sequences of real numbers such that series $\sum a_n^2$ and $\sum b_n^2$ converge. Then the series $\sum a_nb_n$

A. is absolutely convergent

B. may not converge

C.is always convergent, but maynot converge absolutely

D.converges to 0

Attempt

I took $a_n =(-1)^n$ and $b_n=1$. Then , $a_nb_n$ is not convergent.So this removes options C and D.If i take $a_n=\frac{(-1)^n}{n}$ and $b_n=1$, it removes option A. So B is left. But i am not sure

Thanks

For all $a,b \in \mathbb{R}$ we have $$ |ab| = |a| |b| \leq 2 |a| |b| \leq |a|^2 + |b|^2 = a^2 + b^2. $$ Because $\sum_{n=0}^\infty a_n^2 < \infty$ und $\sum_{n=0}^\infty b_n^2 < \infty$ it follows that $$ \sum_{n=0}^\infty |a_n b_n| \leq \sum_{n=0}^\infty (a_n^2 + b_n^2) = \sum_{n=0}^\infty a_n^2 + \sum_{n=0}^\infty b_n^2 < \infty, $$ so the series converges absolutely.

PS: This shows that A. holds, and thus also answers if B. and C. hold. Notice that D. does not hold; take for example $a_0 = b_0 = 1$ and $a_n = b_n = 0$ for $n > 0$.

Hint : Use Cauchy-Schwarz inequality.$$\sum a_nb_n\le \left(\sum a_n^2\right)^{1/2}.\left(\sum b_n^2\right)^{1/2}$$