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Thread: Telescoping series

  1. #1
    Senior Member mollymcf2009's Avatar
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    Telescoping series

    $\displaystyle \sum^{\infty}_{n=1} (e^{\frac{4}{n}} - e^{\frac{4}{n+1}})$

    I'm not sure how to set this up so I can evaluate it as a telescoping sum. Do I need to pull out a negative and make it a ratio? Thanks!
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  2. #2
    MHF Contributor chisigma's Avatar
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    The only you have to do is writing the series ‘term by term’…

    $\displaystyle \sum_{n=1}^{\infty}$$\displaystyle (e^{\frac{4}{n}} - e^{\frac {4}{n+1}}) = e^{4} - e^{\frac{4}{2}} + e^{\frac{4}{2}} - e^{\frac{4}{3}} + e^{\frac{4}{3}} - e^{\frac{4}{4}} + \dots$

    … and the value of the sum is quite evident…

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by chisigma View Post
    The only you have to do is writing the series ‘term by term’…

    $\displaystyle \sum_{n=1}^{\infty}$$\displaystyle (e^{\frac{4}{n}} - e^{\frac {4}{n+1}}) = e^{4} - e^{\frac{4}{2}} + e^{\frac{4}{2}} - e^{\frac{4}{3}} + e^{\frac{4}{3}} - e^{\frac{4}{4}} + \dots$

    … and the value of the sum is quite evident…

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    I get $\displaystyle e^4$ is that correct?
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by mollymcf2009 View Post
    I get $\displaystyle e^4$ is that correct?
    Yes!...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  5. #5
    MHF Contributor chisigma's Avatar
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    Dear Molly…
    … I’m afraid have made a great mistake!… The series you have proposed has termes with alternating sign and decreasing in absolute value. But necessary condition for convergence is that the general term tend to 0 and in this case it isn’t. So the series doesn’t converge… and your thanks to me are totally not deserved

    I apologize and promit never to repeat such a terrible mistake!…

    Very sorry!

    $\displaystyle \chi$ $\displaystyle \sigma$
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  6. #6
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by chisigma View Post
    Dear Molly…
    … I’m afraid have made a great mistake!… The series you have proposed has termes with alternating sign and decreasing in absolute value. But necessary condition for convergence is that the general term tend to 0 and in this case it isn’t. So the series doesn’t converge… and your thanks to me are totally not deserved

    I apologize and promit never to repeat such a terrible mistake!…

    Very sorry!

    $\displaystyle \chi$ $\displaystyle \sigma$
    No worries! I figured it out later after trying it that same way a few times. MEant to post corrections and forgot. Don't sweat it! Thanks for commenting on it though!
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by chisigma View Post
    Dear Molly…
    … I’m afraid have made a great mistake!… The series you have proposed has termes with alternating sign and decreasing in absolute value. But necessary condition for convergence is that the general term tend to 0 and in this case it isn’t. So the series doesn’t converge… and your thanks to me are totally not deserved

    I apologize and promit never to repeat such a terrible mistake!…

    Very sorry!

    $\displaystyle \chi$ $\displaystyle \sigma$


    To evaluate this telescoping series, first find the sum of the first $\displaystyle N$ terms.

    Thus, $\displaystyle \sum_{n=1}^{N}e^{\frac{4}{n}}-e^{\frac{4}{n+1}}=e^{4}-e^{2}+e^2-e^{\frac{4}{3}}+\dots+e^{\frac{4}{N}}-e^{\frac{4}{N+1}}=e^4-e^{\frac{4}{N+1}}$

    Now to evaluate $\displaystyle \sum_{n=1}^{\infty}e^{\frac{4}{n}}-e^{\frac{4}{n+1}}$, take the sum of the first $\displaystyle N$ terms and see what happens as $\displaystyle N\to\infty$.

    In otherwords, evaluate $\displaystyle \lim_{N\to\infty}e^{4}-e^{\frac{4}{N+1}}$
    Thus, $\displaystyle \lim_{N\to\infty}e^{4}-e^{\frac{4}{N+1}}=e^4-e^0=e^4-1$.

    Therefore, $\displaystyle \sum_{n=1}^{\infty}e^{\frac{4}{n}}-e^{\frac{4}{n+1}}=\boxed{e^4-1}$.
    Last edited by Chris L T521; Mar 25th 2009 at 12:04 AM. Reason: small typo.
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  8. #8
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Chris L T521 View Post


    To evaluate this telescoping series, first find the sum of the first $\displaystyle N$ terms.

    Thus, $\displaystyle \sum_{n=1}^{N}e^{\frac{4}{n}}-e^{\frac{4}{n+1}}=e^{4}-e^{2}+e^2-e^{\frac{4}{3}}+\dots+e^{\frac{4}{N}}-e^{\frac{4}{N+1}}=e^4-e^{\frac{4}{N+1}}$

    Now to evaluate $\displaystyle \sum_{n=1}^{\infty}e^{\frac{4}{n}}-e^{\frac{4}{n+1}}$, take the sum of the first $\displaystyle N$ terms and see what happens as $\displaystyle N\to\infty$.

    In otherwords, evaluate $\displaystyle \lim_{N\to\infty}e^{4}-e^{\frac{4}{n+1}}$.

    Thus, $\displaystyle \lim_{N\to\infty}e^{4}-e^{\frac{4}{N+1}}=e^4-e^0=e^4-1$.

    Therefore, $\displaystyle \sum_{n=1}^{\infty}e^{\frac{4}{n}}-e^{\frac{4}{n+1}}=\boxed{e^4-1}$.
    Effectively the general term of the series is…

    $\displaystyle a_{n} = e^{\frac{4}{n}}-e^{\frac{4}{n-1}}$

    … so that is…

    $\displaystyle \lim_{n\to\infty}a_{n}=\lim_{n\to\infty} e^{\frac{4}{n}}\cdot (1-e^{\frac{4}{n+1}}\cdot e^{- \frac{4}{n}})=\lim_{n\to\infty} e^{\frac{4}{n}}\cdot (1-e^{- \frac{4}{n(n+1)}})= 0$

    So the series converges and its sum is what you have computed: $\displaystyle e^{4} -1$. Tonight i discovered to have [probably] done a bad suggestion to Molly and for that i was 'furious' ... This story contains an useful teaching: hurry gives bad suggestions...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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