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agentx5

  • 2 years ago

How would you evaluate this? \[\large \sum_{n=3}^{\infty} \frac{4}{n^2-1}\] o_O I'm thinking there's some trick here I'm not seeing...

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  1. KingGeorge
    • 2 years ago
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    My recommendation would be\[\sum_{n=3}^\infty \frac{4}{n^2-1}=\sum_{n=3}^\infty \frac{4}{(n+1)(n-1)}\]and then apply partial fractions.

  2. agentx5
    • 2 years ago
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    4 = A(n-1) + B(n+1) 4 = An - A + Bn + B A + B = 0 4 = -A + B A + (A+4) = 0 2A = -4 A = -2 B = 2 Now what?

  3. agentx5
    • 2 years ago
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    \[\sum_{n=3}^{\infty} \frac{-2}{n+1} + \sum_{n=3}^{\infty} \frac{2}{n-1}\]

  4. KingGeorge
    • 2 years ago
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    \[\sum_{n=3}^\infty \left(\frac{2}{n-1}-\frac{2}{n+1}\right)\]seems like a telescoping series to me.

  5. agentx5
    • 2 years ago
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    The kind where the middle terms keep cancelling each other out right? Yeah I was just introduced to that today :-D You're saying then all you have left are the first and last terms. Hmm, but how do I manipulate it into that form? I guess it the precise question here. Good call on the partial fractions btw, I didn't see that as an option at first glance

  6. KingGeorge
    • 2 years ago
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    It's already in that form. Just write out the first few terms \[\left(\frac{2}{3-1}-\frac{2}{3+1}\right)+\left(\frac{2}{4-1}-\frac{2}{4+1}\right)+\left(\frac{2}{5-1}-\frac{2}{5+1}\right)+...\]Simplify things a bit, and everything will cancel, and you'll be left with some value and nothing afterwards.

  7. agentx5
    • 2 years ago
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    Ah ok, that makes more sense because http://www.wolframalpha.com/input/?i=telescoping+series&lk=4 wasn't helping :P

  8. KingGeorge
    • 2 years ago
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    In this case, you don't even have to worry about anything at the end since \[\lim_{n\to\infty}\frac{4}{n^2-1}=0\]So the end term would be 0.

  9. agentx5
    • 2 years ago
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    +1000 Kudos! So if we can check if the limit goes to zero first thing off the bat then we're able to confirm convergence? Does that mean the evaluation of the series is also zero?

  10. agentx5
    • 2 years ago
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    Oh wait I think I may have misunderstood, the first few that don't cancel out are what will be left

  11. agentx5
    • 2 years ago
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    Yes?

  12. KingGeorge
    • 2 years ago
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    If the limit goes to 0 it doesn't guarantee convergence of the series. For example, \[\sum_{n=1}^\infty\frac{1}{n}\]does not converge. But the mere fact that it's telescoping with a finite number of non-zero terms means that it converges. And yes, the first few terms that don't cancel out will be the overall sum.

  13. agentx5
    • 2 years ago
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    Wonderful information, I'm evaluating it by hand at the moment. I'll post what I've found when I'm done. :-)

  14. agentx5
    • 2 years ago
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    I got \(\large\frac{5}{3}\) :-)

  15. KingGeorge
    • 2 years ago
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    That looks right.

  16. agentx5
    • 2 years ago
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    1 + 2/3 = 5/3

  17. agentx5
    • 2 years ago
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    Ty for helping out on this one :-D

  18. KingGeorge
    • 2 years ago
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    You're welcome.

  19. klimenkov
    • 2 years ago
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    I'm really sorry to interrupt you. But it is very interesting fact, that \[\sum_{n=3}^{\infty} \frac{4}{n^2-1}=\lim_{n \rightarrow \infty}\left( \frac 53 -\left( \frac{2}{n}+\frac{2}{n-1}\right) \right) \]I enjoy this!

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