Check series convergence using Taylor series Kinda stuck in here too, some kind of help would be greatly appreciated!
Using Taylor expansion check if series :
$$\sum_{n=1}^{\infty}\sqrt{n}(\arctan{(n+1)} - \arctan{n})$$
converges.
Thanks in advance!
 A: Hint
We have $\arctan x+\arctan\frac{1}{x}=\pi/2,\,\forall x>0$ so 
$$\sqrt{n}(\arctan{(n+1)} - \arctan{n})=\sqrt{n}(\arctan\frac{1}{n}-\arctan\frac{1}{n+1})\sim\frac{1}{n^{3/2}},$$
hence we can conclude the convergence of the series by comparaison with the Riemann series.
A: Before we apply Taylor series, we will manipulate a bit. Note that
$$\tan\left(\arctan(n+1)-\arctan n\right)=\frac{1}{1+n(n+1)}.$$
Here we have used the formula for $\tan(u-v)$.
Thus $\arctan(n+1)-\arctan n=\arctan\left(\frac{1}{1+n(n+1)}\right)$.
Now the power series for $\arctan x$ will do the job. For the range we are interested in, it is an alternating series. In particular, we conclude that our difference of arctan's is less than $\frac{1}{1+n(n+1)}$, which is less than $\frac{1}{n^2}$.
A: Hint: Note that the Taylor series of $\arctan(x+1)-\arctan(x)$ at $x=\infty$ is 
$$ \frac{1}{x^2}-\frac{1}{x^3}+\frac{1}{x^5}+O \left( \frac{1}{x^6} \right)$$
$$ \implies \arctan(n+1)-\arctan(n) \sim \frac {1}{n^2},\quad n\to \infty $$
