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What is the value of $$\lim_{X\longrightarrow\infty}\frac{1}{X}\sum^X_{n=1} \left(1 + \frac{1}{n} \right)^n$$ and $$\lim_{X\longrightarrow\infty}\sqrt[X]{\prod^X_{n=1} \left(1 + \frac{1}{n} \right)^n}.$$

Obviously both have a limit since $(1+n^{-1})^n$ becomes $e$ for $X\longrightarrow \infty$ and all previous values are smaller than $e$.

My first idea: What if the limits are $e$ because there are infinite values of the sum/prod that equal $e$ and therefore the first values of the sum, which are smaller than $e$, do not matter.

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    $\begingroup$ See Stolz–Cesàro theorem. $\endgroup$ – Cave Johnson Dec 29 '16 at 12:23
  • $\begingroup$ Mhh doesn't help me $\endgroup$ – lukas.simon Dec 29 '16 at 12:26
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    $\begingroup$ @f.schneider It helps you just put $a_X=\sum_{n=1}^X(1+\frac{1}{n})^n$ and $b_X=X$ $\endgroup$ – kingW3 Dec 29 '16 at 12:37
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  • For the first: $$ \left(1 + \frac{1}{n} \right)^n \xrightarrow[n\to\infty]{} e $$ so by Cesàro ("basic" version, no need for the full Stolz–Cesàro theorem): $$\frac{1}{X}\sum^X_{n=1} \left(1 + \frac{1}{n} \right)^n\xrightarrow[X\to\infty]{} e$$ as well.

  • For the second: $$ \ln \sqrt[X]{\prod^X_{n=1} \left(1 + \frac{1}{n} \right)^n} = \frac{1}{X} \sum^X_{n=1} n \ln \left(1 + \frac{1}{n} \right) $$ and since $n\ln\left(1 + \frac{1}{n} \right) \xrightarrow[n\to\infty]{} 1$, by Cesàro $$ \ln \sqrt[X]{\prod^X_{n=1} \left(1 + \frac{1}{n} \right)^n} \xrightarrow[X\to\infty]{} 1 $$ so that $ \sqrt[X]{\prod^X_{n=1} \left(1 + \frac{1}{n} \right)^n} = \exp\left(\frac{1}{X} \sum^X_{n=1} n \ln \left(1 + \frac{1}{n} \right)\right)\xrightarrow[X\to\infty]{} e^1 = e. $

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