Does series $(\cosh\frac{1}{n} - 1)^p$ converge?

Question

Given: series $\sum^\infty_{n=1}(\cosh\frac{1}{n} - 1)^p$.

Question: Consider different $p$ and find out if it converges.

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I tried to represent $\cosh$ as $\frac{e^x - e^{-x}}{2}$. What I got as $n$-th member: $$\left(\frac{\sqrt[n]{e} + \frac{1}{\sqrt[n]{e}} - 2}{2}\right)^p$$

Notice that degree next to $e$ approaches to $0$ as n approaches to $\infty$:

$$\lim_{n\to\infty} 1/n = 0 \text{ and } \lim_{n\to\infty} -1/n = 0$$

So that I can use Maclaurin series for $e^x$ and get:

$$1 + \frac{1}{n+1} + \frac{1}{2(n+1)^2} + \ldots + \frac{1}{k!(n+1)^k} + \ldots + 1 - \frac{1}{n+1} + \frac{1}{2(n+1)^2} + \ldots + (-1)^k\frac{1}{k!(n+1)^k} + \ldots - 2 = \{\forall k: k \text{ is even}\} \frac{2}{2(n+1)^2} + \ldots + \frac{2}{k!(n+1)^k} + \ldots$$

So put it all together:

$$\left(\frac{1}{2(n+1)^2} + \ldots + \frac{2}{k!(n+1)^k} + \ldots\right)^p \forall k: k \text{ is even}$$

As I know, the sum of convergent series is sum of their sums. But how to prove it strictly for infinite number of series? Induction, started from two series? Is my idea even right?

Answer 1

You can use comparison test. As $$ \lim_{x\to 0}\frac{\cosh x-1}{x^2}=\frac12, $$ the convergence of your series is equivalent to the convergence of $$ \sum_{n=1}^\infty\frac1{n^{2p}} $$

Answer 2

I would directly use the Taylor polynomial for $\cosh x$ and some basic comparisons. We have $$\cosh x-1=\frac{x^2}2+R_2(x)=\frac{x^2}2+O(x^3)$$ We thus have that $(\frac1{2n^2}+cn^{-3})^p\le(\cosh\frac1n-1)^p\le(\frac1{2n^2}+Cn^{-3})^p$ for some constants $c,C$. By comparison with $n^{-2p}$, this is summable if and only if $p>\frac12$.