Does the series $\sum_{n=1}^{\infty} \frac{\tan(2n+1)}{(2+n)^{1/2}}$ converge absolutely?

Question

Does the following series converge absolutely?$$\sum_{n=1}^{\infty} \frac{\tan(2n+1)}{(2+n)^{1/2}}$$

So basically I have a few question similar to this one and I'm not sure if I am proving the (absolute) convergence properly. How do I solve it and similar exercises? Thanks in advance. :)

Answer 1

I will show that $\sum_{n=1}^{\infty} \dfrac{|\tan(2n+1)|}{(2+n)^{1/2}} $ diverges.

$\sum_{n=1}^{\infty} \dfrac{1}{(2+n)^{1/2}} $ diverges so it is enough to show that $\tan(2n+1)$ is very often not small.

Suppose $|\tan(2n+1)|$ is small, say less than $.1$. Then $\tan(2n+3) =\dfrac{\tan(2n+1)+\tan(2)}{1-\tan(2n+1)\tan(2)} \approx\dfrac{\tan(2n+1)-2.185}{1+\tan(2n+1)2.185} $ so $|\tan(2n+3)| \gt \dfrac{1}{1+.2185} \gt 0.82 $.

Therefore $\dfrac{|\tan(2n+1)|}{(2+n)^{1/2}}+\dfrac{|\tan(2n+3)|}{(2+n+1)^{1/2}} \gt \dfrac{0.82}{(3+n)^{1/2}} $

and the sum of these diverges.

In other words, either $\dfrac{|\tan(2n+1)|}{(2+n)^{1/2}} \ge \dfrac{0.1}{(2+n)^{1/2}} $ or $\dfrac{|\tan(2n+1)|}{(2+n)^{1/2}}+\dfrac{|\tan(2n+3)|}{(2+n+1)^{1/2}} \gt \dfrac{0.82}{(3+n)^{1/2}} $ and the sum of either of these diverges.

Answer 2

Consider any $\alpha\in (\frac 32, \frac \pi 2)$ and note that for $k\geq 0$, the interval $(\frac \pi 2 +k\pi -\alpha, \frac \pi 2 +k\pi +\alpha)$ has length $2\alpha\in(3, \pi)$, and thus contains at least one odd integer (and at most two).

Note also that for $x\in (\frac \pi 2 +k\pi -\alpha, \frac \pi 2 +k\pi +\alpha)$, $|\tan(x)|>|\tan(\frac \pi 2 -\alpha)|$.

Each $2n+1$ falls into some $(\frac \pi 2 +k\pi -\alpha, \frac \pi 2 +k\pi +\alpha)$, and in that case $2n+1\leq (k+1)\pi$, hence $n\leq \frac{(k+1)\pi -1}{2}$.

Thus $$\sum_{n=1}^{\infty} \left|\frac{\tan(2n+1)}{(2+n)^{1/2}}\right|\geq \sum_{k=0}^{\infty} \frac{|\tan(\frac \pi 2 -\alpha)|}{\left(2+\frac{(k+1)\pi -1}{2}\right)^{1/2}}$$

The last series obviously diverges (it's of the order $\frac{1}{\sqrt k}$).