Prove that $\sum \tan \frac{n+1}{n^2}$ diverges Prove that 
$$\sum\tan \frac{n+1}{n^2}$$
diverges. 
I know that the solution should be using the limit comparison test with $1/n$. 
Therefore, I think that in the limit:
$$
\lim_{n \to \infty} \frac{\tan \frac{n+1}{n^2}}{1/n}
$$
I should get a positive finite value... but, its not working, I don't know how to solve that limit...
I tried using L'Hospital, but its getting complicated: 
$$
\frac{\frac{1}{\cos^2\frac{n+1}{n^2}}(\frac{n+1}{n^2})'}{-(1/n^2)}
$$
Help.
Thank you. 
 A: Let $L=\lim_{n \rightarrow \infty} \frac{|\tan \frac{n+1}{n^2}|}{\frac{1}{n}}$
Then we have 
$L=\lim_{n \rightarrow \infty} \frac{1}{|\cos\frac{n+1}{n^2}|} \frac{|\sin\frac{n+1}{n^2}|}{\frac{1}{n}}$ 
$=\lim_{n \rightarrow \infty} \frac{1}{|\cos\frac{n+1}{n^2}|}\frac{n+1}{n} \frac{|\sin\frac{n+1}{n^2}|}{\frac{n+1}{n}\frac{1}{n}}$
$=\bigg(|\frac{1}{\cos\lim_{n \rightarrow \infty}\frac{n+1}{n^2}}|\bigg)\bigg(\lim_{n \rightarrow \infty}\frac{n+1}{n}\bigg) \bigg(|\lim_{n \rightarrow \infty} \frac{\sin\frac{n+1}{n^2}}{\frac{n+1}{n^2}}| \bigg)$
$=(1)(1)\bigg(|\lim_{n \rightarrow \infty} \frac{\sin\frac{n+1}{n^2}}{\frac{n+1}{n^2}}| \bigg)$.
Let $x=\frac{n+1}{n^2}.$ Then $x \rightarrow 0$ as $n \rightarrow \infty$. It follows that
$L=|\lim_{x \rightarrow 0} \frac{\sin(x)}{x}|=1.$
This proves that $\sum |\tan \frac{n+1}{n^2}|$ converges. It follows that $\sum \tan \frac{n+1}{n^2}$ converges.
Please let me know if there is any clarification necessary.
A: That is very easy using asymptotic analysis, more precisely equivalence of functions:
We know that a polynomial is asymptotically equivalent to its leading term, hence $$\frac{n+1}{n^2}\sim_\infty \frac n{n^2}=\frac 1n.$$
On the other hand, $\;\tan x\sim_0x$, hence, by substitution,
 $$\tan\frac{n+1}{n^2}\sim_\infty \frac{n+1}{n^2}\sim_\infty\frac 1n.$$
A: In order to do the comparison test, it suffices to let $1/n=x$ and argue that
$$\lim_{n\to\infty}{\tan((n+1)/n^2)\over1/n}=\lim_{x\to0^+}{\tan(x+x^2)\over x}=\lim_{x\to0^+}{(1+2x)\sec^2(x+x^2)\over1}=1$$
using L'Hopital for the second equality.
