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To Evaluate the Limit $$L=\lim_{n \to \infty}\left(1+\sum_{k=1}^{n} \frac{1}{\binom{n}{k}}\right)^n \tag{1}$$

My try:

I tried to use $$\frac{1}{\binom{n}{k}}+\frac{1}{\binom{n}{k+1}}=\frac{n+1}{n} \frac{1}{\binom{n-1}{k}} $$

taking summation both sides from $k=1$ to $k=n$ we get

$$\sum_{k=1}^{n} \frac{1}{\binom{n}{k}}+\sum_{k=1}^{n} \frac{1}{\binom{n}{k+1}}=\frac{n+1}{n} \sum_{k=1}^{n} \frac{1}{\binom{n-1}{k}} \tag{2}$$

Now let $$S=\lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{\binom{n}{k}}$$ we have from $(2)$

$$S+S=S$$

hence $$S=0$$

Now $(1)$ is in form of $1^{\infty}$ Indeterminate form whose limit is given by

$$L=e^\left({\lim_{n \to \infty}}n \times \sum_{k=1}^{n} \frac{1}{\binom{n}{k}}\right)$$

How to proceed now?

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    $\begingroup$ Isn't the quantity in the bracket always $\geq 2$ for any $n \geq 1$? The limit should be infinite. $\endgroup$ – 18cyclotomic Jun 1 '18 at 9:35
  • $\begingroup$ @Umeshshankar Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – gimusi Aug 3 '18 at 21:49
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By Bernoulli's inequality we have that

$$\left(1+\sum_{k=1}^{n} \frac{1}{\binom{n}{k}}\right)^n\ge 1+n\sum_{k=1}^{n} \frac{1}{\binom{n}{k}} \ge 1+n \frac{1}{\binom{n}{n}}=1+n\to \infty$$

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$\left(1+\sum_{k=1}^{n} \frac{1}{\binom{n}{k}}\right)^n \ge \left(1+ \frac{1}{\binom{n}{n}}\right)^n =2^n \to \infty$ as $n \to \infty$.

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