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