The set of all positive values of $a$ for which the series $\displaystyle \sum_{n=1}^{\infty} \left( \frac 1n – \tan^{-1}(\frac 1n) \right) ^{3a}$ converges is
(a) $(0,\frac 13]$
(b) $(0,\frac 13)$
(c) $[\frac 13,\infty)$
(d) $(\frac 13, \infty)$
$\bf{Try}:$
If we choose $a=\frac 13$ then $\left( \frac 1n – \tan^{-1}(\frac 1n) \right) =\frac 1n – \{ \frac 1n – \frac 1{3n^3}+\frac 1{5n^5}-\cdots\}=\frac 1{3n^3}-\frac 1{5n^5}+\cdots$. Which is convergent. Hence option (b) and (d) are false. I'm unable to go further. I also want to know if a direct proof is possible without choosing $a$.
Best Answer
$\frac{1}{n}-tan^{-1}(\frac{1}{n})\approx \frac{1}{3n^3}$for large $n$. Therefore sum converges for $a\gt \frac{1}{9}$