Limit of $\frac{1}{X}\sum^X_n \left(1 + \frac{1}{n} \right)^n$ and$\sqrt[X]{\prod^X_n \left(1 + \frac{1}{n} \right)^n}$?.

Question

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.

Answer 1

• 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. $