If $\sum a_n$ converges, then $\sum \sqrt{a_na_{n+1}}$ converges Prove that if the positve term series $\sum^{\infty}_{n=1}a_n$ is convergent, also $\sum^{\infty}_{n=1}\sqrt{a_na_{n+1}}$ is convergent.
Prove that if the positive term series $\sum^{\infty}_{n=1}a_n$ and $\sum^{\infty}_{n=1}b_n$ are convergent, also $\sum^{\infty}_{n=1}a_nb_n$ is convergent.
I've tried to solve it using comparison test, but no results.

For the first problem, use that AM-GM inequality: $\frac{a_n+a_{n+1}}2\ge\sqrt{a_na_{n+1}}$. Note that $$\sum_{n\ge 1}a_n=\frac{a_1}2+\sum_{n\ge 1}\frac{a_n+a_{n+1}}2\;.$$
For the second problem, first show that $\sum_{n\ge 1}a_n^2$ and $\sum_{n\ge 1}b_n^2$ are convergent and hence that $\sum_{n\ge 1}(a_n^2+b_n^2)$ is convergent. Now apply the AM-GM inequality to note that $$\frac{a_n^2+b_n^2}2\ge\sqrt{a_n^2b_n^2}\;.$$
