# 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.

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 $$\frac{a_n^2+b_n^2}2\ge\sqrt{a_n^2b_n^2}\;.$$
• How do i show that $\sum_{n\ge 1}a_n^2$ is convergent? – 55555 Sep 16 '12 at 7:11
• Since $\sum_{n\ge 1}a_n$ is convergent, its terms approach $0$, so there is an $n_0$ such that $0<a_n<1$ and therefore $0<a_n^2<a_n$ for all $n\ge n_0$. Now use the comparison test. – Brian M. Scott Sep 16 '12 at 7:13
One can also do these problems with Cauchy-Schwarz which states that for $a_n,b_n\geq 0,$ we have $\displaystyle \sum a_n b_n \leq \left(\sum a_n^2 \right)^{1/2} \left(\sum b_n^2 \right)^{1/2}.$ The second problem then follows by the argument in the comments of Brian's answer. It also gives $$\sum \sqrt{a_n a_{n+1} } \leq \left(\sum a_n \right)^{1/2} \left(\sum a_{n+1} \right)^{1/2}$$ which solves the first problem.