- #1

daniel_i_l

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## Homework Statement

Prove that if [tex]0 < \alpha < 1 [/tex] and

[tex]\sqrt[n]{|a_n|} <= 1 - \frac{1}{n^{\alpha}}[/tex] for all n >= 1 then the series

[tex]\sum a_n[/tex] converges.

## Homework Equations

## The Attempt at a Solution

First I did:

[tex]|a_n| <= \left({1 - \frac{1}{n^{\alpha}}}\right)^n =

\left({{1 - \frac{1}{n^{\alpha}}}^{n^{\alpha}}}\right)^{n^{1-\alpha}}[/tex]

and since the limit at infinity of [tex]\left({{1-\frac{1}{n^{\alpha}}}^{n^{\alpha}}}\right)[/tex] is

[tex]\frac{1}{e}[/tex] then there exists an N so that for all n>N

[tex]|a_n| <= \left({\frac{2}{e}}\right)^{n^{1-\alpha}}[/tex]

But how can I show that [tex] \sum \left({\frac{2}{e}}\right)^{n^{1-\alpha}}[/tex]

converges?

Is there a better way to do this?

Thanks.

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