How do I compute this limit? I need answer to the question below $$\lim_{n \rightarrow \infty} \left[\sum_{k=1}^{n}\frac{1}{n \choose k}\right]^n$$ I am not sure of the approach to use in solving this question, I am confused
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2Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Jun 11 '24 at 11:44
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Maybe $e^2$? Numerical evaluation seems to get fairly close to that ... – Matti P. Jun 11 '24 at 11:47
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One can note that $\sum_{k=1}^n \frac1{\binom{n}{k}} = 1 + \frac2{n} + O(\frac1{n^2})$. Therefore, calling $a_n$ the $n$-th term of the sequence I obtain $\left(1+\frac2{n}\right)^n \leq a_n \leq (\left(1+\frac{2+\varepsilon_n}{n}\right))^n$, where $\varepsilon_n$ is positive and tends to $0$. Passing to the limit via the squeeze theorem one gets that the limit is $e^2$.
Massimiliano Foschi
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2This question doesn’t meet the standards for the site. Instead of answering it (which encourages low-quality questions), why not look for a good duplicate target, or help the user by posting comments suggesting improvements? Please also read the meta announcement regarding quality standards. – Anne Bauval Jun 11 '24 at 12:49
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Moreover, your answer (as well as the question) is a duplicate of this more detailed one. – Anne Bauval Jun 11 '24 at 13:10