Evaluate $\lim_{n\to \infty}\prod_{k=2}^{n}\left(1-\frac{1}{k^2}\right)$

Question

How would you solve the following limit?

$$\lim\limits_{n\to \infty}\left(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{n^2}\right)\right)$$

I think that the limit is $0$, because $\frac{1}{n^2}$, and also as $n$ increases the value of $1-\frac{1}{n^2}$ decreases, so the limit is zero.

But how can I show this mathematically, or is the only way I can show this? Thanks in advance for your reply.

As $n$ increases, $n^2$ increases, $\frac 1{n^2}$ decreases, $1-\frac 1{n^2}$ increases (but it's bound above by $1$). — Ben Frankel, Oct 14 '14 at 15:58
Euler did $\prod_{k} (1- \frac{x^2}{k^2})$ http://en.wikipedia.org/wiki/Basel_problem#Euler.27s_approach — orangeskid, Oct 14 '14 at 16:47

mathlove · Accepted Answer · 2014-10-17T23:09:06.107

14

$$\begin{align}\lim_{n\to\infty}\prod_{k=2}^{n}\left(1-\frac{1}{k^2}\right)&=\lim_{n\to\infty}\prod_{k=2}^{n}\frac{(k-1)(k+1)}{k^2}\\&=\lim_{n\to\infty}\frac{\left(\prod_{k=2}^{n}(k-1)\right)\left(\prod_{k=2}^{n}(k+1)\right)}{\prod_{k=2}^{n}k^2}\\&=\lim_{n\to\infty}\frac{(n-1)!\left(\frac{(n+1)!}{2!}\right)}{(n!)^2}\\&=\lim_{n\to\infty}\frac{(n-1)!(n+1)!}{2!\cdot n!\cdot n!}\\&=\lim_{n\to\infty}\frac{n+1}{2n}\\&=\lim_{n\to\infty}\frac{1+\frac 1n}{2}\\&=\frac 12.\end{align}$$

edited Oct 17 '14 at 23:09

answered Oct 14 '14 at 16:00

mathlove

139,939

But why is $\prod_{k=2}^{n}(k+1)$ equal to $(n+1)!$ over 2! ? Where does this 2! derives from ? Thank you. – wonderingdev Oct 14 '14 at 16:15
@JohnG.: $\prod_{k=2}^{n}(k+1)=3\times 4\times\cdots\times n\times (n+1)=\frac{(n+1)!}{2!}$. Can you see why? – mathlove Oct 14 '14 at 16:41
Yes, $1 \times 2$ or 2! isn't included in the series. Thank you ! – wonderingdev Oct 14 '14 at 16:47
@JohnG.: Yes, exactly! You are welcome. – mathlove Oct 14 '14 at 16:49

score 5 · Answer 2 · answered Oct 14 '14 at 15:57

5

Hint

Observe that

$$1-\frac1{n^2}=\frac{\frac{n-1}{n}}{\frac{n}{n+1}}$$

and now telescope using the product.

answered Oct 14 '14 at 15:57

Evaluate $\lim_{n\to \infty}\prod_{k=2}^{n}\left(1-\frac{1}{k^2}\right)$

2 Answers2