How we can show if the sum of $$\lim_{n\rightarrow \infty }\sum_{k=0}^{n} \frac{1}{{n \choose k}}$$ converges and then find the result of the sum if it converges?
Thanks for any help.
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$$ \begin{align} \sum_{k=0}^n\frac1{\binom{n}{k}} &=2+\frac2n+\sum_{k=2}^{n-2}\frac1{\binom{n}{k}} \end{align} $$ However $$ \begin{align} \sum_{k=2}^{n-2}\frac1{\binom{n}{k}} &\le\frac{n-3}{\binom{n}{2}}\\ &=\frac{2(n-3)}{n(n-1)}\\[9pt] &\to0 \end{align} $$ Thus, $$ \lim_{n\to\infty}\sum_{k=0}^n\frac1{\binom{n}{k}}=2 $$
robjohn
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For every $n\geqslant4$, the two extreme terms are $\frac11$, the two terms next to them are $\frac1n$ and each of the remaining $n-3$ terms is at most $\frac2{n(n-1)}$, hence the $n$th sum $S_n$ is such that $$ 2\leqslant S_n\leqslant2+\frac2n+\frac2{n(n-1)}(n-3)\lt2+\frac4n. $$ In particular, $S_n\to2$.
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1we were scooped years ago :-) – robjohn Jun 13 '13 at 21:20