# 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.

• Try to express $\arctan(n + 1) - \arctan n$ in terms of $n$, and see what happens if you expand that. Commented Mar 21, 2013 at 22:22

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.

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 $\frac{1}{n^2}$.

• clever trick with $\tan(u+v)$, haven't seen this before. Commented Mar 21, 2013 at 22:32
• Detail: since $\tan x=\tan y$ is not equivalent to $x=y$, you need a little argument to conclude that $\arctan (n+1)-\arctan n=\arctan(bla)$. You need to invoke that $\arctan (n+1)-\arctan n$ belongs to $(-\pi/2,\pi/2)$. I think... Commented Mar 21, 2013 at 22:43
• Sure. But $\arctan m$ for large $m$ is close to $\pi/2$ but to the left of it, so is clear we are working in a corner of $(0,\pi/2)$ close to $\pi/2$. Commented Mar 21, 2013 at 22:49

Hint: Note that the Taylor series of $\arctan(x+1)-\arctan(x)$ at $x=\infty$ is

$$\frac{1}{x^2}-\frac{1}{x^3}+\frac{1}{x^5}+O \left( \frac{1}{x^6} \right)$$

$$\implies \arctan(n+1)-\arctan(n) \sim \frac {1}{n^2},\quad n\to \infty$$