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

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 :)

• HINT: Take the logarithm. You will find a natural Riemann sum. – Crostul Oct 6 '16 at 11:39

## 2 Answers

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

• Thanks ! So far I have $$ln\left( \prod_{k=1}^{n}\left(1+{k^2\over n^2}\right)^{1\over n}\right)$$ $$= \sum_1^n ln\left(1+{k^2\over n^2}\right)^{1\over n}$$ $$= {1\over n} . \sum_1^n ln\left(1+{k^2\over n^2}\right)$$ Which fits the bill for $a=0 , b=1 , f(x) = ln(1+x^2)$ – Furrane Oct 6 '16 at 11:54

$$\prod_{k=1}^{n}\left(1+\frac{k^2}{n^2}\right)^{1/n} = \exp\left[\frac{1}{n}\sum_{k=1}^{n}\log\left(1+\frac{k^2}{x^2}\right)\right]$$ 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: $$\begin{eqnarray*} \int_{0}^{1}\log(1+x^2)\,dx &=& \left.x\log(1+x^2)\right|_{0}^{1}-\int_{0}^{1}\frac{2x^2}{1+x^2}\,dx\\&=&\log(2)-2+2\int_{0}^{1}\frac{dx}{1+x^2}\\&=&\log(2)-2+2\arctan(1). \end{eqnarray*}$$

• Thanks @Jack D'Aurizio I managed to get to $\int_{0}^{1} log(1+x^2)dx$ but I'm having trouble calculating it's primitive. Can you develop a little on the integration by part ? Thanks in advance – Furrane Oct 6 '16 at 16:33
• @Furrane: all right, I will update the answer in a few minutes. – Jack D'Aurizio Oct 6 '16 at 16:34
• I accepted Zhanxiong because He replied first and gave me the hint I was looking for in the original post, but if I could I would give it to you too since you helped me even further. Thanks a lot =) – Furrane Oct 6 '16 at 16:52
• @Furrane: it is a just trivial algebraic manipulation: $$\frac{2x^2}{1+x^2}=2\cdot\frac{x^2}{1+x^2}=2\cdot\left(1-\frac{1}{1+x^2}\right)$$ – Jack D'Aurizio Oct 7 '16 at 14:36
• now that you say it it seems trivial yes, I've never been good at 'seeing' those fractions modifications. Thanks for the quick answer, you're a savior =) – Furrane Oct 7 '16 at 14:41