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Thread: Limit (4)

  1. #1
    MHF Contributor

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    Limit (4)

    $\displaystyle \lim_{n\to\infty} \sum_{k=0}^n \frac{1}{\binom{n}{k}} = ? $
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  2. #2
    Super Member PaulRS's Avatar
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    Let $\displaystyle
    S_n = \sum_{k=0}^n{\tfrac{1}{\binom{n}{k}}}
    $. If $\displaystyle n$ were odd we can write: $\displaystyle
    S_n = \tfrac{2}
    {{\binom{n}{0}}} + \tfrac{2}
    {{\binom{n}{1}}} + 2\sum\limits_{\tfrac{n}
    {2} \geqslant k \geqslant 2} {\tfrac{1}
    {{\binom{n}{k}}}}
    $ and if it were even: $\displaystyle
    S_n = \tfrac{2}
    {{\binom{n}{0}}} + \tfrac{2}
    {{\binom{n}{1}}} + 2\sum\limits_{\tfrac{n}
    {2} - 1 \geqslant k \geqslant 2} {\tfrac{1}
    {{\binom{n}{k}}}} + \tfrac{1}
    {{\binom{n}{n/2}}}
    $

    In any case: $\displaystyle 0 \leq
    S_n-2 \leq \tfrac{2}
    {{\binom{n}{1}}} + 2\sum\limits_{\tfrac{n}
    {2} \geqslant k \geqslant 2} {\tfrac{1}
    {{\binom{n}{k}}}}
    $ , since $\displaystyle \binom{n}{k}$ is unimodal we may write: $\displaystyle 0\leq\sum\limits_{\tfrac{n}
    {2} \geqslant k \geqslant 2} {\tfrac{1}
    {{\binom{n}{k}}}} \leq{(n/2)\cdot \tfrac{1}{\binom{n}{2}}}\to{0^+}$, hence $\displaystyle \sum\limits_{\tfrac{n}
    {2} \geqslant k \geqslant 2} {\tfrac{1}
    {{\binom{n}{k}}}}\to{0}$ and so, by the Sandwich Theorem: $\displaystyle
    \lim_{n\to{+\infty}}{(S_n - 2)}=0
    $ i.e. $\displaystyle
    \lim_{n\to{+\infty}}{S_n}=2
    $
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  3. #3
    Senior Member DeMath's Avatar
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    We can also use this $\displaystyle \left( {\begin{array}{*{20}{c}}n \\k \\\end{array} } \right) \geqslant \left( {\begin{array}{*{20}{c}}n \\2 \\\end{array} } \right)$ if $\displaystyle 2 \leqslant k \leqslant n - 2$.
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