Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.


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

I can't see anything in this limit , so help me please.

share|improve this question

5 Answers 5

up vote 25 down vote accepted

Note that $$1-\frac1{k^2}=\left(1-\frac1k\right)\left(1+\frac1k\right)=\frac{k-1}{k}\frac{k+1}{k}=\frac{a_k}{a_{k-1}}$$ with $a_k= \frac{k+1}k$, hence this is a telescoping product, i.e. $$ \prod_{k=2}^n\left(1-\frac1{k^2}\right)=\frac{a_2}{a_1}\frac{a_3}{a_2}\cdots\frac{a_n}{a_{n-1}}=\frac{a_n}{a_1}=\frac{n+1}{2n}.$$

share|improve this answer
I think that is \frac{a_n}{a_1} –  Sherloek holmes Dec 26 '12 at 19:24

Let $g(k)=\frac{k-1}{k}$. Then this product is:

$$\prod_{k=2}^n \frac{g(k)}{g(k+1)}$$

which is a telescoping product, and thus equal to $\frac{g(2)}{g(n+1)}$

share|improve this answer

$$ 1-\frac{1}{k^2} = \frac{(k-1)(k+1)}{k^2}$$ $$ \prod_{k=2}^{n}\left ( 1-\frac{1}{k^2} \right ) = \underbrace{\frac{2-1}{2}}_{\text{from first}} \times \underbrace{\frac{(n+1)}{n}}_{\text{from last}}$$

share|improve this answer

Hint: The "typical" term is $\dfrac{k-1}{k}\dfrac{k+1}{k}$. Express the first few terms in this way, and observe the nice cancellations.

For example, here is the product of the first seven terms: $$\frac{1}{2}\frac{3}{2}\frac{2}{3}\frac{4}{3}\frac{3}{4}\frac{5}{4}\frac{4}{5}\frac{6}{5}\frac{5}{6}\frac{7}{6}\frac{6}{7}\frac{8}{7}\frac{7}{8}\frac{9}{8}=\frac{9}{2\cdot 8}.$$

share|improve this answer



Observe that $\left(1-\frac1{(k)^2}\right)$ is cancelled out by the previous & the next term except for the extreme terms, the 1st & the last term leaving behind the 1st part of the 1st term $=\frac{2-1}{2}$ and the 2nd part of the last term $=\frac{n+1}n$

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.