If $\sum_{n=0}^\infty a_n$ converges and $a_n>0$, does $\sum_{n=0}^\infty \sqrt{a_na_{n+1}}$ converge?

Question

If the sum: $$\sum_{n=0}^\infty a_n$$ converges and $a_n>0$, does it mean that: $$\sum_{n=0}^\infty \sqrt{a_na_{n+1}}$$ converges too?

Answer 1

Yes: compare to $\frac12\sum_n(a_n+a_{n+1})$.

Answer 2

It is true tha converges.

Use the inequality: $$ ab \leqslant \frac{a^2}{2}+\frac{b^2}{2}$$ for $a=\sqrt{a_n}$ and $b=\sqrt{a_{n+1}}$

As a second way you can use the Cauchy-Swarz inequality:

$\sum_{n=1}^{\infty}\sqrt{a_na_{n+1}} \leqslant \sum_{n=1}^{\infty}\sqrt{a_n}^2 \sum_{n=1}^{\infty}\sqrt{a_{n+1}}^2$

Answer 3

Note: $\sqrt{a_n \cdot a_{n+1}} \le a_n+a_{n+1}$.

$\sum_{n=0}^{\infty} \sqrt{a_n \cdot a_{n+1}} \le 2\sum_{n=0}^{\infty} a_n$