Convergent or divergent series: $\sum_{n=1}^{\infty} \sqrt{a_n \cdot a_{n+1}}$?

Question

suppose that $\sum_{n=1}^{\infty}a_n$ is a convergent series of positive terms. Prove that $\sum_{n=1}^{\infty} \sqrt{a_n \cdot a_{n+1}}$ is also convergent. Demonstrate that the converse is false.

Answer 1

Note that $2\sqrt{a_n a_{n+1}}\le a_n+a_{n+1}$. (This is a disguised version of $(x-y)^2\ge 0$.) Then use Comparison.

For the falsity of the converse, let $a_n=1$ if $n$ is odd, and let $a_n$ be a fairly rapidly convergent series when $n$ is even. For example, $\frac{1}{n^4}$ will work.

Answer 2

By the arithmeticg-geometric inequality $\sqrt{xy}\le \frac{x+y}2$ for $x,y\ge0$. Therefore the second series is dominated as $$\sum_{n=1}^\infty\sqrt{a_na_{n+1}}\le\sum_{n=1}^\infty\frac{a_n+a_{n+1}}2=\frac12\left(\sum_{n=1}^\infty a_n+\sum_{n=2}^\infty a_{n+1}\right)=-\frac{a_1}2+\sum_{n=1}^\infty a_n.$$

The converse is false as can be seen from $a_{2n}=n^{-4}$, $a_{2n+1}=1$.