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

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

Michael Hardy
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GorillaApe
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  • See Basel problem. – Lucian Feb 24 '14 at 19:17
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    @Lucian Probably not needed. – Sawarnik Feb 24 '14 at 20:23
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    @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 '14 at 02:21

4 Answers4

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.

André Nicolas
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12

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 $$a_n=\frac{\displaystyle\prod_{k=2}^nv_k}{\displaystyle\prod_{k=2}^nv_{k+1}}=\frac{\displaystyle\prod_{k=2}^nv_k}{\displaystyle\prod_{k=3}^{n+1}v_{k}}=\frac{v_2}{v_{n+1}}=v_2\times\frac{n+1}{n}\to v_2=\frac12$$

6

Answer:

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

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 $= 1/2$.

Michael Hardy
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Satish Ramanathan
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  • how do you get the first row ? – GorillaApe Feb 24 '14 at 20:19
  • @Parhs Write each term in the form $(k+1)/k\cdot(k-1)/k$. Get it? – Sawarnik Feb 24 '14 at 20:22
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    @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 '14 at 21:02
  • yep got it. How did you figure it out ? I am trying to find an explanation -am I retarded. – GorillaApe Feb 24 '14 at 21:50
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    @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 '14 at 22:19
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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.

Cameron Williams
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