TITLE: Series of Geometric Means Converges
QUESTION [11 upvotes]: I'm trying to prove the following:
Let $\{b_{n}\}_{n \ge 1}$ be a sequence of strictly positive terms such that
$$\sum_{n = 1}^{\infty}b_{n}$$
converges. Set $\gamma_{n} = (b_{1} \cdots b_{n})^{1/n}$. Prove
$$\sum_{n = 1}^{\infty}\gamma_{n}$$
converges.
Attempt: Letting $b = \sum b_{n}$, and using AM-GM:
\begin{align*}
            \sum_{n = 1}^{\infty}(b_{1} \cdots b_{n})^{\frac{1}{n}} &< \sum_{n = 1}^{\infty}b_{1}\cdots b_{n} \\
            &< \sum_{n = 1}^{\infty}\frac{1}{n^{n}}(b_{1}+\cdots+b_{n})^{n} \\
            &< \sum_{n = 1}^{\infty}\left(\frac{b}{n}\right)^{n}
        \end{align*}
 This last series converges by the root test, so $\sum\gamma_{n}$ converges by comparison, given $b_{1} \cdots b_{n} \ge 1$ for all $n$. If $b_{1} \cdots b_{n} < 1$ for finite $n$, then we can reduce this case to the previous argument by adding a sufficiently large constant to $\sum(\frac{b}{n})^{n}$. My problem is that I dont know how to deal with the case: $b_{1} \cdots b_{n} < 1$ for infinite $n$...

REPLY [2 votes]: You can also prove this using Hardy's inequality:
Set $a_n=\sqrt b_n.$ Then we have $\sum_{n=1}^\infty b_n = \sum_{n=1}^\infty a_n^2 <\infty.$
Now write 
\begin{align}
\sum_{n=1}^\infty \gamma_n &= \sum_{n=1}^\infty ((a_1\cdots a_n)^{1/n})^2 \\
&\le \sum_{n=1}^\infty \left(\frac{1}{n}\sum_{k=1}^n a_k\right)^2 \\
&< 4\sum_{n=1}^\infty a_n^2 < \infty.
\end{align}