$\displaystyle \lim_{n\to\infty} \sum_{k=0}^n \frac{1}{\binom{n}{k}} = ?$
2. 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. $\displaystyle \lim_{n\to{+\infty}}{S_n}=2$
3. 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 $\displaystyle 2 \leqslant k \leqslant n - 2$.