Convergence of series $\displaystyle\sum\frac{1}{\sqrt n}\tan\left(\frac1{n}\right)$ How to test convergence of:
$$\displaystyle\sum\frac{1}{\sqrt n}\tan\left(\frac1{n}\right)$$
I tried to use Raabe's Test but it did not work.
In logarithmic test I find it difficult to calculate the limit.
I cannot proceed further.
Kindly help me.
 A: Compare with harmonic $\sum \frac{1}{n} $. We have
$$\frac{ \frac{1}{\sqrt{n}} \tan (1/ \sqrt{n} ) }{\frac{1}{n}} = \sqrt{n} \tan (1/ \sqrt{n}) = \frac{ \tan(1/ \sqrt{n} )}{\frac{1}{\sqrt{n}}} \to 1$$
Why? Let $t = \frac{1}{\sqrt{n}}$. Then, as $n \to \infty$, $t \to 0$. So
$$ \lim_{n \to \infty} \frac{ \tan(1/ \sqrt{n} )}{\frac{1}{\sqrt{n}}} = \lim_{t \to 0} \frac{ \tan t}{t} = 1 $$
Hence, your series diverges!
Update:
The question has been changed, now we have $\sum \frac{1}{ \sqrt{n}} \tan (1/n) $. In this case, we claim the series converges. Indeed, compare with the convergent series $\sum \frac{1}{n^{3/2} }$. Then, we have
$$ \frac{ \frac{1}{\sqrt{n}} \tan(1/n) }{ \frac{1}{n^{3/2} }} =n \tan(1/n) = \frac{ \tan(1/n) }{1/n} =_{t = 1/n} \frac{ \tan t}{t} \to 1 \; \text{as} \; t \to 0 $$
and thus the series converges!
A: If $0<x<\dfrac \pi 2$ then $\tan x >x$.  Therefore
$$
\sum_{n=1}^\infty \frac 1 {\sqrt n} \tan \frac 1 {\sqrt n} \ge \sum_{n=1}^\infty \frac 1 {\sqrt n} \cdot \frac 1 {\sqrt n} = \sum_{n=1}^\infty \frac 1 n =\infty.
$$
And now the problem has been altered by a later edit. Let's look at the new version.  Observe that if $0<x<c$ for $c$ small enough, then $\tan x < 2x$.  Therefore
$$
\sum_n \frac 1 {\sqrt n} \tan \frac 1 n \le \sum_n \frac 1 {\sqrt n}\cdot \frac 2 n = 2 \sum_n \frac 1 {n^{3/2}} < \infty.
$$
A: Using Raabe's test with $$u_n=\frac{1}{\sqrt n}\tan\left(\frac1{n}\right)$$ $$\frac{u_n}{u_{n+1}}=\frac{\sqrt{n+1} \tan\left(\frac{1}{n}\right)}{\sqrt{n} \tan\left(\frac{1}{n+1}\right)}$$ Now, using Taylor expansion for large values of $n$ $$\frac{u_n}{u_{n+1}}=1+\frac{3}{2 n}+\frac{3}{8 n^2}+O\left(\frac{1}{n^{5/2}}\right)$$ $$n\left(\frac{u_n}{u_{n+1}}-1\right)=\frac{3}{2}+\frac{3}{8 n}+O\left(\frac{1}{n^{3/2}}\right)>1$$ then the series will be absolutely convergent. 
A: This is a special case of: If $f$ is any function on $[0,1]$ such that $f(0)=0$ and $f'(0)$ exists, then
$$\tag 1  \sum_{n=1}^{\infty}\left |\frac{f(1/n)}{\sqrt n}\right | < \infty.$$
Examples: $f(x) = \tan x, \ln (1+x),  \dots.$
Proof of $(1)$: For $h>0,$ $h$ small, we have
$$\left |\frac{f(h)}{h} \right | = \left|\frac{f(h)-f(0)}{h}\right | \le 2|f'(0)|.$$
Thus for large $n$ we have $|f(1/n)| \le 2|f'(0)|(1/n).$ Therefore the terms of the series in $(1)$ are $\le 2|f'(0)|/n^{3/2}$ for large $n.$ Since $\sum 1/n^{3/2}< \infty,$ $(1)$ follows by the comparison test. 
A: let $u = \frac{1}{\sqrt n}$
$\displaystyle\sum u \tan u $
Since taylor series for $u \tan u$ is $u^2 + \frac{u^4}{3} + \frac{2u^6}{15}+O(u^8)$
So $\displaystyle\sum u \tan u  = \sum u^2 + \frac{u^4}{3} + \frac{2u^6}{15}+O(u^8) = \sum \frac{1}{n} + \frac{1}{3n^2} + \frac{2}{15n^3} + O(\frac{1}{n^4}) \geq   \sum \frac{1}{n} = +\infty $
So the summation divergents. 
A: zeta regularization the series converge sorry to contradict you
$$\sum _{j=1}^{\infty } \left(\frac{\tan \left(\frac{1}{j}\right)}{\sqrt{j}}=\sum _{j=1}^{\infty } \left(\frac{\tan \left(\frac{1}{j}\right)}{\sqrt{j}}-\left(\frac{1}{j}\right)^{3/2}\right)+\zeta \left(\frac{3}{2}\right)=3.21574\right.$$
