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$$\sum_{k=1}^n\frac{1}{\binom{n}{k}} = 1+\frac{2}{n}+\sum_{k=2}^{n-2}\frac{1}{\binom{n}{k}}\Rightarrow 1+\frac{2}{n}\leq\sum_{k=1}^n\frac{1}{\binom{n}{k}}$$ $$\sum_{k=1}^n\frac{1}{\binom{n}{k}}\leq 1+\frac{2}{n}+\frac{4}{n(n-1)}+\frac{6(n-5)}{n(n-1)(n-2)} = 1+\frac{2}{n}+\frac{10n-38}{n(n-1)(n-2)}=:f(n)$$ \begin{align*} \log\left(\lim_{n\to\infty}f(n)^n\right) & = \lim_{n\to\infty}\log(f(n)^n)\\ &= \lim_{n\to\infty}n\log(f(n))\\ & = \lim_{n\to\infty}\frac{\log(f(n))}{f(n)-1}(f(n)-1)n\\ & = \lim_{n\to\infty}\frac{\log(f(n))}{f(n)-1}\lim_{n\to\infty}(f(n)-1)n\\ & = \lim_{n\to\infty}(f(n)-1)n\\ \end{align*} Hence, $$\lim_{n\to\infty}f(n)^n = \exp\left(\lim_{n\to\infty}(f(n)-1)n\right) = e^2$$
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Sat, 02 Oct 2021 11:18 GMT