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I have to study the convergence of the following series: $$\sum_{n=0}^{\infty}\cos{\left(\frac{n+2}{n^2+4}\right)}$$ I have thought to apply the asymptotic criterion.
I know that $\cos{\left(\frac{n+2}{n^2+4}\right)}\sim 1-\left(\frac{n+2}{n^2+4}\right)^2\cdot \frac{1}{2}$.
Then it is right to say that since: $$1-\left(\frac{n+2}{n^2+4}\right)^2\cdot \frac{1}{2}\sim -\frac{1}{2n^2}\,\,\ (*)$$ then the original series converges?

I am not sure it is possible to"concatenate" the asymptotic equivalence as I have done in (*)

$\color{red}{Note:}$ by $a_n\sim b_n$ I intend that the $\lim_{n\to\infty}\frac{a_n}{b_n}=1$

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    $\begingroup$ That is too much work for this problem. What is the limit of the sequence being summed? $\endgroup$ Jul 2, 2021 at 16:33
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    $\begingroup$ The terms approach $1$, so it definitely diverges. $\endgroup$
    – K.defaoite
    Jul 2, 2021 at 16:39
  • $\begingroup$ I have no idea where (*) comes from, but it's definitely wrong. For example, as $n\to\infty$ the LHS tends to $1$ while the RHS tends to $0$. $\endgroup$ Jul 2, 2021 at 17:38

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A necessary condition for convergence is that the general term tends to zero. However $$\frac{n+2}{n^2+4}=\frac{1/n+2/n^2}{1+4/n^2}\to0$$ and therefore $$\cos\left(\frac{n+2}{n^2+4}\right)\to1$$

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