# Homework Help: Telescoping Series

1. Oct 7, 2006

### courtrigrad

$$\sum_{n=1}^{\infty} \arctan(n+1)-\arctan(n)$$.

So we want to take $$\lim s_{n} = \lim_{n\rightarrow \infty} (\arctan 2-\arctan 1)+(\arctan 3-\arctan 2) + ... + (\arctan(n+1)-\arctan n)$$. From this I can see all of the terms cancel except $$\arctan 1$$. But then how do we get: $$\lim_{n\rightarrow \infty} \arctan(n+1)-\arctan 1$$? Wouldn't the $$\arctan(n+1)$$ cancel out?

Thanks

2. Oct 7, 2006

### courtrigrad

I got it. I was assuming that it would keep on going. But it stops. I have to take the limit. so $$\arctan(n+1)$$ and a $$-\arctan 1$$ are left.

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