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Homework Help: Telescoping Series

  1. Oct 7, 2006 #1
    [tex] \sum_{n=1}^{\infty} \arctan(n+1)-\arctan(n) [/tex].

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

  2. jcsd
  3. Oct 7, 2006 #2
    I got it. I was assuming that it would keep on going. But it stops. I have to take the limit. so [tex] \arctan(n+1) [/tex] and a [tex] -\arctan 1 [/tex] are left.
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