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Thread: [SOLVED] prove that this series to the nth^2 power converges

  1. #1
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    [SOLVED] prove that this series to the nth^2 power converges

    Prove that

    $\displaystyle \sum\left(1-\frac{1}{n}\right)^{n^2}$

    converges.
    My instinct is to use the root test, and try to show that

    $\displaystyle \lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n<1$.

    But unfortunately I don't know how to show this either.

    Are there any tricks I can use on these nth root proofs which might help here?

    Thanks!
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Note that for each $\displaystyle n\ge1$ is $\displaystyle \left( 1-\frac{1}{n} \right)^{n}\le \frac{1}{e},$ and then $\displaystyle \left( 1-\frac{1}{n} \right)^{n^{2}}\le e^{-n},$ besides the series $\displaystyle \sum_{n\ge1}e^{-n},$ converges by the integral test, thus the original series does converge.

    ----

    Of course your idea does work too. In order to prove that $\displaystyle \left( 1-\frac{1}{n} \right)^{n}\to \frac{1}{e}$ as $\displaystyle n\to\infty,$ we have $\displaystyle 1-\frac{1}{n}=\frac{n-1}{n}=\frac{1}{\frac{n}{n-1}}=\frac{1}{1+\frac{1}{n-1}}\implies \left( 1-\frac{1}{n} \right)^{n}=\frac{1}{\left( 1+\frac{1}{n-1} \right)^{n}},$

    so it's easy to see that $\displaystyle \left( 1+\frac{1}{n-1} \right)^{n}\to e$ as $\displaystyle n\to\infty,$ and hence, the result.
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  3. #3
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    Thank you very much, but I could still benefit from further assistance.

    Once we have $\displaystyle \left( 1-\frac{1}{n} \right)^{n}\le \frac{1}{e}$, the proof is easy for me. However, I am uncertain how you arrived at that inequality.

    We have by definition

    $\displaystyle e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$.

    I have no idea how to prove that this series is increasing, but let's say for a moment that I could. In that case, we have, for each $\displaystyle m=n-1$,

    $\displaystyle e>\left(1+\frac{1}{m}\right)^m$.

    Then algebraically we can show that

    $\displaystyle e>\left(\frac{m+1}{m}\right)^m$

    flipping:

    $\displaystyle \frac{1}{e}<\left(\frac{m}{m+1}\right)^m$.

    Substitute back $\displaystyle m=n-1$ to get

    $\displaystyle \frac{1}{e}<\left(\frac{n-1}{n}\right)^{n-1}$,

    and expand:

    $\displaystyle \frac{1}{e}<\left(1-\frac{1}{n}\right)^{n-1}$.

    But this isn't good enough to get to our goal, which, again, is:

    $\displaystyle \left( 1-\frac{1}{n} \right)^{n}\le \frac{1}{e}$.

    Apparently I need to learn more about exponents. Any more tricks would be much appreciated.
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