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From an old Putnam : Prove that if $(x_n)$ is a sequence of positive real numbers, and $\sum{x_n}$ converges, then so does $\sum{(x_n)^{\frac{n}{n+1}}}$.

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You can divide the terms into two classes : one decreasing sufficiently fast compared to a geometric sequence, and the others. Give separate estimates for each case. –  sos440 Mar 18 '13 at 23:25
    
good problem ,good hint..+1 –  Halil Duru Mar 19 '13 at 0:17

1 Answer 1

up vote 2 down vote accepted

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

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