Discuss whether the series $\sum \left[(\pi/2)^a - (\arctan n)^a\right]$ converges or not, based on the value of $a$

Question

$$\sum_{n=1}^\infty {\left[ {{{\left( {\frac{\pi }{2}} \right)}^a} - {{(\arctan n)}^a}} \right]} $$ I proved that the series diverges for $a < 0 $ and that the series converges for $a = 1$ (using $\arctan n + \arctan 1/n = \pi/2$). But I can't discuss whether it converges or not for $a>0$.

Answer 1

If $x>y>0$ and $a>0$, $$\frac{x^a-y^a}{x-y} = a\, z^{a-1},\quad z\in(y,x) $$ by Lagrange's theorem, hence by taking $x=\frac{\pi}{2}$ and $y=\arctan n$ we have that the convergence $$ \sum_{n\geq 1}\left(\left(\frac{\pi}{2}\right)^a-\left(\arctan n\right)^a\right)$$ just depends on the convergence of: $$ \sum_{n\geq 1}\left(\frac{\pi}{2}-\arctan n\right) = \sum_{n\geq 1}\arctan\frac{1}{n}.$$ However, the last series is diverging by asymptotic comparison with the harmonic series, so the original series is never converging for $a>0$.