# Find limit of this decreasing sequence

$$a_n=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right) \cdots \left(1-\frac{1}{n^2}\right)$$

I have proved that this sequence is decreasing. However I am trying to figure out how to find its limit.

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See Basel problem. –  Lucian Feb 24 at 19:17
@Lucian Probably not needed. –  Sawarnik Feb 24 at 20:23
@Sawarnik: Couldn't agree more! (I don't recall saying otherwise...) However, I'd like to think that the people who post here and visit this site are genuinely interested in expanding their mathematical horizon, as opposed to merely getting a quick fix for whatever homework they were given... Maybe I'm wrong. But at least I present them with an opportunity: Whether they take it or not is up to them; at least I know I did my duty. –  Lucian Feb 25 at 2:21

Hint: Rewrite each $1-\frac{1}{k^2}$ as $\frac{(k-1)(k+1)}{k^2}$ and observe the mass cancellations. It will be useful to do this explicitly for say the product of the first $5$ terms.

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+1 @'mass cancellation' –  r9m Feb 24 at 17:38
Hehe..."serial mass cancelator"...+1 –  DonAntonio Feb 24 at 17:45
Products of mass cancellation –  robjohn Feb 24 at 19:42

We write

$$a_n=\prod_{k=2}^n\left(1-\frac{1}{k^2}\right)=\prod_{k=2}^n\frac{(k-1)(k+1)}{k^2}=\prod_{k=2}^n\frac{v_k}{v_{k+1}}$$ where $$v_k=\frac{k-1}{k}$$ so by change of index $$a_n=\frac{\displaystyle\prod_{k=2}^nv_k}{\displaystyle\prod_{k=2}^nv_{k+1}}=\frac{\displaystyle\prod_{k=2}^nv_k}{\displaystyle\prod_{k=3}^{n+1}v_{k}}=\frac{v_2}{v_{n+1}}=v_2\times\frac{n+1}{n}\to v_2=\frac12$$

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$$\left(\frac{3}{2}\frac{4}{3}\frac{5}{4}\cdots\frac{n}{n-1}\frac{n+1}{n}\right)\cdot\left(\frac{1}{2}\frac{2}{3}\frac{3}{4}\cdots\frac{n-2}{n-1}\frac{n-1}{n}\right)$$

After mass cancellations, pull the $$\frac{n+1}{2}\text{ and }\frac{1}{n}$$

$$\frac{n+1}{2}\cdot\frac{1}{n}$$

Limit of this function tending to infinity $= 1/2$.

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how do you get the first row ? –  Parhs Feb 24 at 20:19
@Parhs Write each term in the form $(k+1)/k\cdot(k-1)/k$. Get it? –  Sawarnik Feb 24 at 20:22
@sawarnik, Thanks for answering. this is exactly was Andre was suggesting but I made it explicitly laid out so that it is clear –  satish ramanathan Feb 24 at 21:02
yep got it. How did you figure it out ? I am trying to find an explanation -am I retarded. –  Parhs Feb 24 at 21:50
@Parhs, If you asked this question with three votes, you are intelligent. Further, this question is being answered by heavyweights in this site. The language sometimes is very difficult to assimilate by students. I used to be like you when I started out. No worry. Keep learning –  satish ramanathan Feb 24 at 22:19

Hint: try defining $b_n = \ln(a_n)$ (which is well-defined) and see what limit this goes to. Then use a certain exponential function to see what $\lim_n a_n$ is.

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explain why the log is a powerful tool? products become sums, etc. –  PatrickT Feb 25 at 12:38