1. ## Limit (4)

$\lim_{n\to\infty} \sum_{k=0}^n \frac{1}{\binom{n}{k}} = ?$

2. Let $
S_n = \sum_{k=0}^n{\tfrac{1}{\binom{n}{k}}}
$
. If $n$ were odd we can write: $
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: $
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: $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 $\binom{n}{k}$ is unimodal we may write: $0\leq\sum\limits_{\tfrac{n}
{2} \geqslant k \geqslant 2} {\tfrac{1}
{{\binom{n}{k}}}} \leq{(n/2)\cdot \tfrac{1}{\binom{n}{2}}}\to{0^+}$
, hence $\sum\limits_{\tfrac{n}
{2} \geqslant k \geqslant 2} {\tfrac{1}
{{\binom{n}{k}}}}\to{0}$
and so, by the Sandwich Theorem: $
\lim_{n\to{+\infty}}{(S_n - 2)}=0
$
i.e. $
\lim_{n\to{+\infty}}{S_n}=2
$

3. 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 $2 \leqslant k \leqslant n - 2$.