Riemann Sums - $ \lim_{n\to +\infty} \prod_{k=1}^{n}\left(1+{k^2\over n^2}\right)^{1\over n} = ? $

Question

This is about a homework I have to do. I don't want the straight answer, just the hint that may help me start on this. To give you context, we worked on series, and we're now studying integrals, linking the two with Riemann sums.

Now here is the question :

Using Riemann sums, find :

$$ \lim_{n\to +\infty} \prod_{k=1}^{n}\left(1+{k^2\over n^2}\right)^{1\over n} $$

Which is to say, find $a,b\in \mathbb{R}$, $f$ a function, so that :

$$ \lim_{n\to +\infty} \prod_{k=1}^{n}\left(1+{k^2\over n^2}\right)^{1\over n} = \left(b-a\over n\right)\sum_{k=1}^n f\left(a+k.{b-a\over n}\right) $$

I didn't find any way to simplify the "Pi" product.

Thanks to anyone kind enough to help me :)

Answer 1

Hint: take $\log$ first then make the Riemann sum.

Answer 2

$$\prod_{k=1}^{n}\left(1+\frac{k^2}{n^2}\right)^{1/n} = \exp\left[\frac{1}{n}\sum_{k=1}^{n}\log\left(1+\frac{k^2}{x^2}\right)\right]$$ hence, by Riemann sums and integration by parts: $$ \lim_{n\to +\infty}\prod_{k=1}^{n}\left(1+\frac{k^2}{n^2}\right)^{1/n} = \exp\left[\int_{0}^{1}\log(1+x^2)\,dx\right]=\color{red}{2\cdot e^{\pi/2-2}}.$$ Explanation: $$\begin{eqnarray*} \int_{0}^{1}\log(1+x^2)\,dx &=& \left.x\log(1+x^2)\right|_{0}^{1}-\int_{0}^{1}\frac{2x^2}{1+x^2}\,dx\\&=&\log(2)-2+2\int_{0}^{1}\frac{dx}{1+x^2}\\&=&\log(2)-2+2\arctan(1). \end{eqnarray*}$$