I need to prove if the series converges or diverges:


I'm sure it converges, because each of in $n\rightarrow \infty $ $a_{n}$ is 0. But I'm having trouble proving it.

I need to do it without the limit comparison test (checking if the limit is more than 1, or ..).

Any help/hints will be appreciated.

  • $\begingroup$ No. I mean when $x\rightarrow\infty$ They are 1. so the total is 0. $\endgroup$ – sheldonzy May 19 '17 at 15:27
  • $\begingroup$ "I'm sure it converges, because each of in $n \to \infty$ $a_n$ is $0$." If you're saying that it converges because $\lim_{n \to \infty} a_n = 0$, this does not guarantee convergence; e.g. $a_n = \frac{1}{n}$. $\endgroup$ – TastyRomeo May 19 '17 at 15:34

Maybe you are interested in $\sum_{n=1}^{\infty}(\cosh(1/n)-\cos(1/n))$ (with plus sign it is trivially divergent because the term goes to $2\not=0$.

Hint. Consider the Taylor expansion of $\cosh(x)-\cos(x)$ at $0$: $$\cosh(x)-\cos(x)=\frac{1}{2}(e^x+e^{-x}-2\cos(x))\\=\frac{1}{2}(1+x+\frac{x^2}{2}+1-x+\frac{x^2}{2}-2(1-\frac{x^2}{2}))+o(x^2)=x^2+o(x^2).$$ What may we conclude?

P.S. Alternatively find $\lim_{x\to 0}\frac{\cosh(x)-\cos(x)}{x^2}$ by using Hopital.

  • $\begingroup$ Thanks. Unfortunately I haven't learned Taylor expansions yet. Any other direction? $\endgroup$ – sheldonzy May 19 '17 at 15:32
  • 2
    $\begingroup$ Find the limit of $(\cosh(x)-\cos(x))/x^2$ as $x\to 0$ by using Hopital. $\endgroup$ – Robert Z May 19 '17 at 15:35

As already remarked, $\cosh\frac{1}{n} = 1+\frac{1}{2n^2}+O\left(\frac{1}{n^4}\right)$ and $\cos\frac{1}{n}=1-\frac{1}{2n^2}+O\left(\frac{1}{n^4}\right)$, hence their difference behaves like $\frac{1}{n^2}$ for large values of $n$, and $\sum_{n\geq 1}\frac{1}{n^2}$ is a convergent series. Using the full Taylor series,

$$ \sum_{n\geq 1}\left(\cosh\frac{1}{n}-\cos\frac{1}{n}\right)=\sum_{m\geq 0}\frac{2\,\zeta(4m+2)}{(4m+2)!}\approx \color{red}{1.64776}. $$


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