TITLE: Convergence of this series
QUESTION [10 upvotes]: From an old Putnam : Prove that if $(x_n)$ is a sequence of positive real numbers, and $\sum{x_n}$ converges, then so does $\sum{(x_n)^{\frac{n}{n+1}}}$.

REPLY [2 votes]: Following sos440's hint.
For every $x > 0$ and $n \in \Bbb N$, we have
$$
x^\frac{n}{n+1} \leq \max\left(2^{-n},2x\right)< 2x + 2^{-n}
$$
Indeed:

either $x \leq 2^{-(n+1)}$ and $x^\frac{n}{n+1} \leq 2^{-n}$,
or $x > 2^{-(n+1)}$ and $x^\frac{n}{n+1} = x \times x^{-\frac{1}{n+1}} < 2x$

So,
$$
\sum_{n=0}^\infty x_n^\frac{n}{n+1} < 2\sum_{n=0}^\infty x_n + \sum_{n=0}^\infty 2^{-n} < \infty
$$