Get ready Part 8

Doing this intriguing problem hastily on a Sunday morning, I came up with an awkward "piecewise" solution. There must be something more elegant.

Let $b_n=a_n^{\frac{n}{n+1}}$ and note that $\frac{a_n}{b_n}=\sqrt[n+1]{a_n}$ . Considering two cases, we will show that $b_n< \frac{1}{2^n}+2a_n$ for all $n$ , so that $\sum b_n$ converges by comparison.

Indeed, if $a_n\leq \frac{1}{2^{n+1}}$ then $b_n\leq \frac{1}{2^n}$ and if $a_n>\frac{1}{2^{n+1}}$ then $\frac{a_n}{b_n}=\sqrt[n+1]{a_n}> \frac{1}{2}$ so $b_n<2a_n$ .

Let be $\sum_{n=1}^{\infty} a_{n}$ a convergent serie and $\lbrace a_n \rbrace$ its corresponding sequence.

Then let be the sequence $\lbrace a_{n}^{\frac{n}{n+1}} \rbrace$ ; take n $\in \mathbb{N}$ then as $\frac{n}{n+1} < 1$ .

And as $a_n^x$ is a non-decreasing function for any a $\in \mathbb{R^+}$ you got that

$a_n^{\frac{n}{n+1}} < a_n^1$

Then take the sum of the first n terms of both sequences and take the limit; finally you got that as $\sum_{n=1}^{\infty} a_{n}$ is convergent the other also is