If $a_n\ge0$ and $\sum a_n$ converges then $\sum\sqrt{a_na_{n-1}}$ converges, what about the converse?
Solution 1:
No, it is not true. Define $a_n = n^2$ for $n$ even and $a_n = n^{-100}$ for $n$ odd. Then clearly $\sum a_n$ is divergent, but each term
$$\sqrt{a_n a_{n - 1}} \sim n^{-49}$$
gives a convergent series.
Solution 2:
The converse is not ture. For example consider the following sequense $\{a_n\}$. $$a_n = \begin{cases} \dfrac{1}{2^nn} & \text{if $n$ is even} \\[2ex] n & \text{if $n$ is odd .} \end{cases}$$
Then $\sum\limits_{n=1}^{\infty}\sqrt{a_na_{n-1}}$ is convergent. But $\sum\limits_{n=1}^{\infty}{a_n}$ is divergent.