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How to do this limit?

  1. Nov 14, 2004 #1
    Hi all, I tried to do this limit, but didn't find a way:

    \lim_{n \rightarrow \infty} \left( 1 - \frac{1}{2^2} \right) \left( 1 - \frac{1}{3^2} \right) ... \left( 1 - \frac{1}{n^2} \right)

    I tried to use theorems I know so far, but it didn't lead to success. Will somebody help how to do these kinds of limits (I know there is no general rule, just some advice what should I try when I'm asked to find limits of such sequences)

    The same case is with:

    \lim_{n \rightarrow \infty} \left( \frac{1}{n} - \frac{2}{n} + \frac{3}{n} - ... + \frac{(-1)^{n-1}n}{n} \right)


    \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{k(k + 1)}

    I hope I will be able to go on with a small hint....

    Thank you.
  2. jcsd
  3. Nov 14, 2004 #2


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    For the first limit, try writing the kth term as [tex]\frac{k^2-1}{k^2}[/tex]. Factor the numerator (difference of squares). Now write out the first 4 or 5 terms of your product in this form. Notice anything nice?

    For the second, consider the sum [tex]S_n=1-2+3-...+(-1)^{n-1}n[/tex] Can you find a general form for[tex]S_n[/tex]? Write out the first few terms if need be. Hint:what is it when n is odd? even? Try writing n=2k or n=2k+1 respectively for n even/odd. You should be a step closer to your limit now.

    For your third series, try partial fractions on the terms. Write out the first few terms and what do you see?
  4. Nov 14, 2004 #3


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    Dearly Missed

    For your first:
    This ought to be a rather suggestive form...

    Arrgh..I was beaten.
    Last edited: Nov 14, 2004
  5. Nov 16, 2004 #4
    That's great, now I can guess the limit is 3/4. Thank you!
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