On convergence of the series. $\sum_{n=1}^{\infty}\left[\frac{1}{n}-\tan^{-1}\left(\frac{1}{n}\right)\right]^{a}$ The question is to find set of all positive values of a for which the series $$\displaystyle\sum_{n=1}^{\infty}\left[\frac{1}{n}-\tan^{-1}\left(\frac{1}{n}\right)\right]^{a}$$ converges. Looking at $a_n $ I am just confused how to proceed.Which test I should try?
 A: Note that:
$$tan^{-1}\left(\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{3n^3}+o\left(\frac{1}{n^3}\right)$$
thus
$$\frac{1}{n}-tan^{-1}\left(\frac{1}{n}\right)=\frac{1}{3n^3}+o\left(\frac{1}{n^3}\right)$$
therefore the series converges for $$a>\frac13$$
A: We have
$${1\over n}-\arctan\left(1\over n\right)=\int_0^{1/n}\left(1-{1\over1+x^2}\right)dx=\int_0^{1/n}{x^2\over1+x^2}dx$$
and, for $n\ge1$, we have
$${1\over6n^3}=\int_0^{1/n}{x^2\over2}dx\le\int_0^{1/n}{x^2\over1+x^2}dx\le\int_0^{1/n}x^2dx={1\over3n^3}$$
so
$${1\over6}\sum_{n=1}^\infty{1\over n^{3a}}\le\sum_{n=1}^\infty\left({1\over n}-\arctan\left(1\over n\right)\right)^a\le{1\over3}\sum_{n=1}^\infty{1\over n^{3a}}$$
Thus by the well-known comparison test, the series in question converges if and only if $a\gt{1\over3}$.
A: From the expansion
$$\frac {1}{1+x^2}=1-x^2+x^2\epsilon (x) $$
we get
$$\arctan (x)=x-\frac {x^3}{3}(1+\epsilon (x)) $$
and
$$\frac {1}{n}-\arctan(\frac {1}{n})\sim \frac {1}{3n^3} $$
thus
$$u_n\sim \frac {1}{3^an^{3a}} $$
It converges if $$a>\frac {1}{3} $$
A: Since at $x=0$ we have  $\arctan(x)\sim x-x^3/3+o(x^3)~~$ then,
$$\lim_{n\to \infty}\frac{1/n-\arctan(1/n)}{1/n^3}=\lim_{x\to 0}\frac{x-\arctan(x)}{x^3}= \frac{1}{3}$$
that is 
$$\left(\frac1n-\arctan(\frac1n)\right)^a\sim \frac{1}{n^{3a}}$$

Whence by Riemann series, the convergence holds if and only if $3a>1$.

