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Math Help - limit of sequence

  1. #1
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    limit of sequence

    I need to show that the limit of the sequence,

    \frac{{n + 1}}{{{n^2}}} + \frac{{{{(n + 1)}^2}}}{{{n^3}}} +  \cdots  + \frac{{{{(n + 1)}^n}}}{{{n^{n + 1}}}}

    is e-1. What's sad is I have the solution in front of me and I can't understand the simple algebra.

    \frac{{n + 1}}{{{n^2}}}\left[ {\frac{{1 - {{\left( {\frac{{n + 1}}{n}} \right)}^n}}}{{1 - \frac{{n + 1}}{n}}}} \right] = \frac{{n + 1}}{n}\left[ {{{\left( {1 + \frac{1}{n}} \right)}^n} - 1} \right]

    The second factor on the left, how did they get that?
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  2. #2
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    Re: limit of sequence

    Quote Originally Posted by zg12 View Post
    I need to show that the limit of the sequence,

    \frac{{n + 1}}{{{n^2}}} + \frac{{{{(n + 1)}^2}}}{{{n^3}}} +  \cdots  + \frac{{{{(n + 1)}^n}}}{{{n^{n + 1}}}}

    is e-1. What's sad is I have the solution in front of me and I can't understand the simple algebra.

    \frac{{n + 1}}{{{n^2}}}\left[ {\frac{{1 - {{\left( {\frac{{n + 1}}{n}} \right)}^n}}}{{1 - \frac{{n + 1}}{n}}}} \right] = \frac{{n + 1}}{n}\left[ {{{\left( {1 + \frac{1}{n}} \right)}^n} - 1} \right]

    The second factor on the left, how did they get that?
    One "n" in the denominator of the left factor was taken inside the brackets.

    \frac{{n + 1}}{{{n^2}}}\left[ {\frac{{1 - {{\left( {\frac{{n + 1}}{n}} \right)}^n}}}{{1 - \frac{{n + 1}}{n}}}} \right] = \frac{n+1}{n}\left[\frac{1-\left(\frac{n+1}{n}\right)^n}{n\left(1-\frac{n+1}{n}\right)}\right]

    = \frac{n+1}{n}\left[\frac{1-\left(\frac{n+1}{n}\right)^n}{n\left(\frac{n-n-1}{n}\right)}\right]

    and the rest follows.

    Do you understand how the first term and common ratio of the geometric series,
    were placed into the formula for summing the series ?

    I think I misread your question a little.

    a=u_1=\frac{n+1}{n^2}

    r=\frac{n+1}{n}

    S_n=\frac{a\left(1-r^n\right)}{1-r}

    It is of course a "series", rather than a sequence.
    Last edited by Archie Meade; June 29th 2011 at 02:53 AM.
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  3. #3
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    Re: limit of sequence

    Quote Originally Posted by Archie Meade View Post
    Do you understand how the first term and common ratio of the geometric series,
    were placed into the formula for summing the series ?
    Oh goodness, I am a moron. Thanks, I see what happened. I guess they assumed that would have been at the surface of my mind from when I was in high school. not!
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  4. #4
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    Re: limit of sequence

    It may have been offputting to see "n" as the term in the numerator and denominator
    as it is also the number of terms in the series.
    Last edited by Archie Meade; June 29th 2011 at 02:55 AM.
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