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Find the limit $$\lim_{n \to \infty}\left[\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{n^2}\right)\right]$$

I take log and get $$\lim_{n \to \infty}\sum_{k=2}^{n} \log\left(1-\frac{1}{k^2}\right)$$ How can I solve.Please help.

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See also the near-duplicate: math.stackexchange.com/q/18179 –  Martin Feb 16 '13 at 5:53

3 Answers 3

up vote 21 down vote accepted

The identity $k^2-1=(k+1)(k-1)$ shows that $$ \prod_{k=2}^n\left(1-\frac{1}{k^2}\right)=\prod_{k=2}^n\frac{k^2-1}{k^2}=\prod_{k=2}^n\frac{k-1}k\cdot\prod_{k=2}^n\frac{k+1}k=\frac1n\cdot\frac{n+1}2, $$ and the value of limit should follow.

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We can call it a "telescoping product". –  GEdgar Jan 25 '13 at 18:45
    
@GEdgar Absolutely. One probably applies telescopic more often to sums but this fits perfectly products as well. –  Did Jan 25 '13 at 20:07
    
@Did: this looks very close to lab bhattacharjee's answwer, which was first. –  robjohn Feb 15 '13 at 20:02
    
@robjohn Thanks for your comment. I am impressed by the fact that (as a moderator?) you would see fit to mention the precedence by 21 seconds of (the first version of) an answer over this one. –  Did Feb 15 '13 at 20:53

The $r$th term is $\frac{r^2-1}{r^2}=\frac{(r-1)}r\frac{(r+1)}r$

So, the product of $n$ terms is $$\frac{3.1}{2^2}\frac{4.2}{3^2}\frac{5.3}{4^2}\cdots \frac{(n-1)(n-3)}{(n-2)^2}\frac{(n-2)n}{(n-1)^2}\frac{(n-1)(n+1)}{n^2}$$ $$=\frac12\frac32\frac23\frac43\cdots\frac{n-2}{n-1}\frac n{n-1}\frac{n-1}n\frac{n+1}n=\frac12\frac{(n+1)}n$$

as the 1st half of any term is cancelled by the last half of the previous term except for the 1st & the last term.

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Your way works nice if you employ the Euler's infinite product for the sine function. Then

$$\lim_{n \to \infty}\sum_{k=2}^{n} \ln\left(1-\frac{1}{k^2}\right)=\lim_{x\to\pi}\ln\left(\frac{\pi^2\sin x}{x(\pi^2-x^2)}\right)=\lim_{y\to0}\ln\left(\frac{\pi^2\sin y}{y(2\pi-y)(\pi-y)}\right)=\ln(1/2)$$ Thus, your product is $1/2$.

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To Chris,s sister would you like to explain me Infinite product of sine function and how can i write $\displaystyle \lim_{n \to \infty}\sum_{k=2}^{n} \log\left(1-\frac{1}{k^2}\right)=\lim_{x\to\pi}\ln\left(\frac{\pi^2\sin x}{x(\pi^2-x^2)}\right)$ Thanks –  juantheron Jan 25 '13 at 18:58
    
@juantheron: I simply used Euler's infinite product for the sine function, took log of both sides and then rearranged things to get the expression you see after the first equal sign. Then it remained to take the limit to $\pi$ to get our limit. For more details on infinite product of sine function, see here: ams.org/bookstore/pspdf/gsm-97-prev.pdf –  Chris's sis Jan 25 '13 at 19:38
    
Thanks Chris's sister –  juantheron Jan 25 '13 at 21:04
    
@juantheron: welcome! –  Chris's sis Jan 25 '13 at 21:06

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