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Calculate the limit of the sequence

$$\lim_{n\rightarrow\infty}\ a_n$$

$$a_n=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\ldots\left(1-\frac{1}{n^2}\right), n\geqslant2 $$

Here is what I did:

$\left(1-\frac{1}{n^2}\right)=\left[\left(1-\frac{1}{n^2}\right)^{n^2}\right]^\frac{1}{n^2}=e^\frac{1}{n^2}$

$a_n=e^{\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}}$

$\Rightarrow \lim_{n\rightarrow\infty}\ a_n=e^{\frac{1}{\infty}}=e^0=1$

Not sure if I'm on the right track..

Thank you very much in advance! Really great to see what an amazingly smart community this website has.

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Apply $\ln$ to $a_n$ -- do you know the limit of what you get to see? –  user20266 May 25 '12 at 18:04
    
Great idea, I'll try that right now. Thank you! –  Grozav Alex Ioan May 25 '12 at 18:07

1 Answer 1

up vote 4 down vote accepted

Hint: We have $$1-\frac{1}{k^2}=\frac{k^2-1}{k^2}=\frac{k-1}{k}\frac{k+1}{k}.\tag{$1$} $$ Write out the product of the first few terms, in the expanded form described in $(1)$. There will be a whole lot of collapsing (cancellation) going on.

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Wow, now that was simple. Thanks a lot! Solved it :)! –  Grozav Alex Ioan May 25 '12 at 18:22

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