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

Did
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A.D
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3 Answers3

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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.

Did
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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.

lab bhattacharjee
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    Like https://math.stackexchange.com/questions/2245151/evaluating-limit-of-an-infinite-product/2245164#2245164, $$\prod_{r=2}^n\left(1-\dfrac1{r^2}\right)=\prod_{r=2}^n\dfrac{T_{r-1}}{T_r}=\dfrac{T_1}{T_n}$$ where $$T_m=\dfrac m{m+1}$$ – lab bhattacharjee Apr 25 '17 at 15:21
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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 $1/2$.

user 1591719
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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: http://www.ams.org/bookstore/pspdf/gsm-97-prev.pdf – user 1591719 Jan 25 '13 at 19:38