Does $\sum_{n=1}^\infty \frac {\arctan\,n }{n^{\frac65}}$ converge or diverge?

Question

Reason behind question

These were my thoughts on how to prove it.

Since $\arctan\,n \gt \dfrac1n$ for $n\gt2$, we can say $\dfrac {\arctan\,n }{n^{\frac65}} \gt \dfrac 1{n^{\frac{11}5}}$

Now we use limit comparison test, and since bn converges, and we get a number as a result of our limit, $\dfrac {\arctan\,n }{n^{\frac65}}$. Does this converge?

Answer 1

Comparison test: $\tan^{-1}(n)<\pi/2$ for all $n$, hence

$$ \sum_{n=1}^\infty \frac{\tan^{-1}(n)}{n^{1.2}} < \frac{\pi}{2} \sum_{n=1}^\infty \frac{1}{n^{1.2}} $$

and the latter sum converges (the sum over $n^{-\alpha}$ converges for all $\alpha>1$).

Answer 2

As $\arctan$ is bounded, we can compare $\sum_n\frac{\arctan n}{n^{1.2}}$ with $\sum_n\frac1{n^1.2}$, which converges because $1.2>1$.