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Math Help - Limit (4)

  1. #1
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    Limit (4)

    \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 <br />
S_n = \sum_{k=0}^n{\tfrac{1}{\binom{n}{k}}}<br />
. If n were odd we can write: <br />
S_n  = \tfrac{2}<br />
{{\binom{n}{0}}} + \tfrac{2}<br />
{{\binom{n}{1}}} + 2\sum\limits_{\tfrac{n}<br />
{2} \geqslant k \geqslant 2} {\tfrac{1}<br />
{{\binom{n}{k}}}} <br />
and if it were even: <br />
S_n  = \tfrac{2}<br />
{{\binom{n}{0}}} + \tfrac{2}<br />
{{\binom{n}{1}}} + 2\sum\limits_{\tfrac{n}<br />
{2} - 1 \geqslant k \geqslant 2} {\tfrac{1}<br />
{{\binom{n}{k}}}}  + \tfrac{1}<br />
{{\binom{n}{n/2}}}<br />

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