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.
Best Answer
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.