# Study the series $\sum_{n=0}^{\infty}\cos{\left(\frac{n+2}{n^2+4}\right)}$

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 $$\lim_{n\to\infty}\frac{a_n}{b_n}=1$$

• That is too much work for this problem. What is the limit of the sequence being summed? Jul 2, 2021 at 16:33
• The terms approach $1$, so it definitely diverges. Jul 2, 2021 at 16:39
• 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 $0$. Jul 2, 2021 at 17:38

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 $$\cos\left(\frac{n+2}{n^2+4}\right)\to1$$