A limit with a product and number e

Question

How can we proof this fact? $$\lim_{n\to\infty} \sqrt[n] {\displaystyle\prod_{k=1}^{n} e^{- \left (\frac kn \right )^2}}= \frac {1}{\sqrt[3] e} .$$

Answer 1

It is easy to notice, using simple properties of exponentiation that $$\prod_{i=1}^na^{x_i}=a^{x_1}\cdots a^{x_n}=a^{x_1+\cdots+x_n}=a^{\sum\limits_{i=1}^nx_i}$$ For the sake of understandable text, I am going to be denoting $e^{f(x)}$ as $\exp(f(x))$ for the rest of this solution.

Considering the function inside the limit, $$\sqrt[n]{\prod_{k=1}^ne^{-\left(\frac{k}{n}\right)^2}}$$ We can first use properties of exponents to write this as: $$\prod_{k=1}^ne^{-\frac{k^2}{n^2}\times\frac{1}{n}}=\prod_{k=1}^ne^{-\frac{k^2}{n^3}}$$ Now, we use the property above to rewrite this as, $$\exp\left(\sum_{k=1}^n\frac{-k^2}{n^3}\right)$$ $$\exp\left(-\frac{1}{n^3}\sum_{k=1}^nk^2\right)$$ Read How to get to the formula for the sum of squares of first n numbers? to know how we rewrite this as the following: $$\exp\left(-\frac{1}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}\right)$$ $$\exp\left(-\frac{2n^3+3n^2+n}{6n^3}\right)$$ $$\exp\left(-\left(\frac{1}{3}+\frac{1}{2n}+\frac{1}{6n^2}\right)\right)$$ As $n\to\infty$, $2n\to\infty$ and $6n^2\to\infty$ too. Therefore, $\frac{1}{2n}\to 0$ and $\frac{1}{6n^2}\to 0$ So, $$\lim_{n\to\infty}\exp\left(-\left(\frac{1}{3}+\frac{1}{2n}+\frac{1}{6n^2}\right)\right)$$ Plugging in values, $$\exp\left(-\frac{1}{3}\right)$$ $$e^{-\frac{1}{3}}$$ $$\frac{1}{\sqrt[3]{e}}$$ and we are done.

Answer 2

Often converting a product to a sum helps out.

And a formula: $\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}$

$y=(\Pi_{k=1}^n e^{-k^2/n^2})^{1/n}$

$\ln y = (1/n) \sum_{k=1}^n\frac{-k^2}{n^2}=\frac{-n(n+1)(2n+1)}{6n^3}$

$\lim_{n\to \infty} \ln y = -1/3$

$\lim_{n \to \infty} y = e^{-1/3}$