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Math Help - Limits

  1. #1
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    Limits

    Help me please with 2 questions:
    Find limit of
    1. a_{n}=\frac{1}{n-1}+...+\frac{1}{2n}

    2. \underset{n \to \infty }{\lim} \left [ \frac{1}{n^{2}}+...+\frac{1}{(2n)^{2}} \right ]


    Thanks!
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Limits

    What have you tried?

    1. n\to ?
    Last edited by Siron; December 17th 2011 at 03:54 AM.
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  3. #3
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    Re: Limits

    Quote Originally Posted by Siron View Post
    What have you tried?

    1. n\to ?

    2. If n\to \infty then \frac{1}{n^2}\to 0, same for the other terms.

    1. n\to \infty
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Limits

    The limit of a sum is the sum of the limits of the separated terms.
    Last edited by Siron; December 17th 2011 at 03:54 AM.
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  5. #5
    MHF Contributor chisigma's Avatar
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    Re: Limits

    Quote Originally Posted by sinichko View Post
    Find limit of
    1. a_{n}=\frac{1}{n-1}+...+\frac{1}{2n}

    Thanks!
    Considering that is...

    \lim_{n \rightarrow \infty} \{\sum_{k=1}^{n-2} \frac{1}{k} - \ln (n-2) \} =\gamma (1)

    \lim_{n \rightarrow \infty} \{\sum_{k=1}^{2n} \frac{1}{k} - \ln 2 n \} =\gamma (2)

    ... subtracting (1) from (2) You obtain...

    \lim_{n \rightarrow \infty} \sum_{k=n-1}^{2n} \frac{1}{k} = \lim_{n \rightarrow \infty} \{\ln 2 n - \ln (n-2)\}= \ln 2 (3)



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  6. #6
    Member sbhatnagar's Avatar
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    Re: Limits

    You could also try another approach:

    \lim_{n \to \infty} \left[  \frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}+\cdots +\frac{1}{2n}\right]
    \lim_{n \to \infty} \left[  \frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}+\cdots +\frac{1}{2n}\right]

    = \left[ \lim_{n \to \infty}\frac{1}{n-1}+\lim_{n \to \infty}\frac{1}{n}+ \lim_{n \to \infty}\sum_{r=1}^{n}\frac{1}{n+r}\right]

    = \left[  0+0+\lim_{n \to \infty}\sum_{r=1}^{n}\left(\frac{1}{n} \right)\frac{1}{1+{r \over n}}\right]

    =\int_{0}^{1} \frac{1}{1+x} dx

    =[\ln(1+x)]^1_0

    =\ln(2)

    \lim_{n \to \infty} \left[  \frac{1}{n^2}+\cdots +\frac{1}{(2n)^2}\right]
    \lim_{n \to \infty} \left[  \frac{1}{n^2}+\cdots +\frac{1}{(2n)^2}\right]=0
    Last edited by sbhatnagar; December 16th 2011 at 09:28 PM.
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  7. #7
    MHF Contributor chisigma's Avatar
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    Re: Limits

    Quote Originally Posted by sinichko View Post
    2. \underset{n \to \infty }{\lim} \left [ \frac{1}{n^{2}}+...+\frac{1}{(2n)^{2}} \right ]

    Defining a_{n}= \frac{1}{n^{2}}+...+\frac{1}{(2 n)^{2}} it is evident that 0<a_{n}<\frac{1}{n} so that is \lim_{n \rightarrow \infty} a_{n}=0...



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  8. #8
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Limits

    Quote Originally Posted by Siron View Post
    What have you tried?



    2. If n\to \infty then \frac{1}{n^2}\to 0, same for the other terms.
    Whrong!
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  9. #9
    MHF Contributor Siron's Avatar
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    Re: Limits

    Quote Originally Posted by Also sprach Zarathustra View Post
    Whrong!
    You're right.
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  10. #10
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    Re: Limits

    need some help in this please!!
    is given the function {y=sin2x/x for x different from 0} or {y=2a for x=o}. find "a" if this function is continuous everywhere.
    please help
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