For which a this series converge? $\sum\limits_{n=1}n^{a+1}\int_n^{n+\log(n)}\frac{\arctan(t^a)}{1+t^{2a}}$ For which $a$ $\in$ $\mathbb{R}$ this series converge?
$$\sum_{n=1}^\infty{n^{a+1}\int_n^{n+\log(n)}\frac{\arctan(t^a)}{1+t^{2a}}}$$
 A: Case 1. If $a > 0$, then
$$ \frac{\arctan (t^{a})}{1 + t^{2a}} \asymp \frac{1}{t^{2a}} \quad \text{as } t \to \infty. $$
Similarly, integrating, we find that
$$ \int_{n}^{n+\log n} \frac{\arctan (t^{a})}{1 + t^{2a}} \, dt \asymp \int_{n}^{n+\log n} \frac{dt}{t^{2a}} \asymp \frac{\log n}{n^{2a}}. $$
Therefore the series converges if and only if
$$ \sum_{n=1}^{\infty} \frac{\log n}{n^{a-1}} $$
converges, which is true if and only if $a > 2$.
Case 2. If $a = 0$, then clearly the series diverges.
Case 3. If $a < 0$, let $b = -a > 0$ for simplicity. Then
$$ \frac{\arctan (t^{a})}{1 + t^{2a}} = \frac{\arctan (t^{-b})}{1 + t^{-2b}} \asymp \frac{1}{t^{b}} \quad \text{as } t \to \infty. $$
Then it follows that
$$ \int_{n}^{n+\log n} \frac{\arctan (t^{a})}{1 + t^{2a}} \, dt \asymp \int_{n}^{n+\log n} \frac{dt}{t^{b}} \asymp \frac{\log n}{n^{b}}, $$
and the series converges if and only if
$$ \sum_{n=1}^{\infty} \frac{\log n}{n^{2b-1}} $$
converges, which is true if and only if $b > 1$.
Conclusion. Therefore, the series converges if and only if $a < -1$ or $a > 2$.
