# Help with series math problem

1. Dec 24, 2004

I'm just not so sure on how to approach this problem. Well, here it goes:

$$\sum _{n=1} ^{\infty} \left[ \tan ^{-1} (n+1) - \tan ^{-1} (n) \right] = \frac{\pi}{2}$$

I know that

$$\tan ^{-1} x = \sum _{n=0} ^{\infty} \left( -1 \right) ^n \frac{x^{2n+1}}{2n+1}$$

but I don't know if it can be useful to get to the answer above. I just need some tips. Any help is highly appreciated.

Last edited: Dec 24, 2004
2. Dec 24, 2004

### Hurkyl

Staff Emeritus
Try writing out the first few terms.

P.S. I get $\pi/4$.

3. Dec 24, 2004

Oh... I see. By the way, you're right about the $\pi/4$.