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Here is a question in calculus. Compute the limit of the sequence: $\lim\limits_{n\rightarrow+\infty}(\sum\limits_{i=1}^n(1+\frac{i}{n})^i)^{\frac{1}{n}}$?

There are in general three ways to compute the limit of a sequence of the form in summation:

(1) Compute the sum directly, and then treat it as the limit of a function. However, it seems that it is impossible to work out what $(\sum\limits_{i=1}^n(1+\frac{i}{n})^i)^{\frac{1}{n}}$ is.

(2) Apply squeeze theorem. I think that this is the most possible way to solve this question, but I cannot find a suitable upper bound and a suitable lower bound.

(3) Regard it as a Riemann sum, and then compute the Riemann integral. In this way, we usually need a factor $\frac{1}{n}$ as the length of the intervals in the partition. Hence, it seems that we need to take logarithm and obtain $\frac{1}{n}\ln\sum\limits_{i=1}^n(1+\frac{i}{n})^i$. However, $\ln$ does not commute with $\sum$, so I do not think that this method will work.

pie
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Hebe
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  • @M.Wind Wolfram Alpha says that this converges to $1$. – Sai Mehta Oct 29 '23 at 17:09
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    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 recognise 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. – Shaun Oct 29 '23 at 19:07
  • I think this can be done by Cauchy limit theorem which mean the sum is equal to $ n e$ then $(ne)^{\frac{1}{n}} \to 1$ – pie Oct 30 '23 at 04:28
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    @pie Note that the last term of the sum is $2^n$, whence the limit is at least $2$ (which it actually is). – Gary Oct 30 '23 at 04:33
  • @Gary you are right – pie Oct 30 '23 at 04:35
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    What about $$\lim_{n\to\infty}n\left(\left[\sum_{k=1}^n\left(1+\frac kn\right)^k\right]^{1/n}-2\right)?$$ – metamorphy Oct 30 '23 at 04:50
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    @metamorphy It seems to be $$
    • 2\log \left( {1 - \frac{1}{{2\sqrt {\rm e} }}} \right) = 0.7227012296175 \ldots

    $$

     – Gary Oct 30 '23 at 06:41

1 Answers1

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$$ \text{as @Gary pointed out in the comments $ \ \ \ $} \sum_{i=1} ^n \left( 1+ \frac{i}{n} \right)^i \geq 2^n $$

notice that $2^n$ is the biggest term in the sequence $\left( 1+ \frac{i}{n} \right)^i$ for $i \in \{1,2,3 \dots ,n \}$ then we can do the following $$ 2^n \leq \sum_{i=1} ^n \left( 1+ \frac{i}{n} \right)^i \leq n2^n$$ $$ 2 \leq \left( \sum_{i=1} ^n \left( 1+ \frac{i}{n} \right)^i \right)^{\frac{1}{n}} \leq 2 \sqrt[n]{n}$$ since $\lim\limits_{n \to \infty } \sqrt[n]{n} =1 $ then by squeeze theorem $$\lim\limits_{n \to \infty } \left( \sum_{i=1} ^n \left( 1+ \frac{i}{n} \right)^i \right)^{\frac{1}{n}} =2$$

pie
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