trasdulaive

2022-02-28

I have the following series:

$\sum _{n=1}^{\mathrm{\infty}}({2}^{\frac{1}{n}}+{2}^{-\frac{1}{n}}-2)$

ImpudgeIntemnect

Beginner2022-03-01Added 6 answers

Note that ${e}^{x}+{e}^{-x}-2={x}^{2}+\frac{{x}^{4}}{12}+\frac{{x}^{6}}{360}+\dots +\frac{2{x}^{2n}}{\left(2n\right)!}+\dots$

It's easy to show that this series has a bound of the form$e}^{x}+{e}^{-x}-2<k{x}^{2$ , where k is some positive constant, for all x sufficiently small. We could use the Taylor remainder theorem or, more simply, just compare the series to the clearly greater geometric series $x}^{2}+{x}^{4}+{x}^{6}+{x}^{8}+\dots =\frac{{x}^{2}}{1-{x}^{2}$ , which gives us the bound (for example)

$e}^{x}+{e}^{-x}-2<\frac{4}{3}{x}^{2$ valid for $\left|x\right|<\frac{1}{2}$

Your sum is just adding the values of this expression at$x=\frac{\mathrm{ln}2}{n}$ , so it has the same convergence properties as $\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{n}^{2}}$ , which converges.

It's easy to show that this series has a bound of the form

Your sum is just adding the values of this expression at

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