Problem of the Week #105 - March 31st, 2014

[Chris L T521]

Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Find the sum of the series $\displaystyle \sum_{n=2}^{\infty}\ln\left(1-\frac{1}{n^2}\right)$

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by Aryth, chisigma, lfdahl, magneto, MarkFL, Opalg, and Pranav. You can find Aryth's detailed solution below.

[sp]We have to find the limit of \(\displaystyle \sum_{n=2}^{\infty}ln\left(1 - \frac{1}{n^2}\right)\).

Claim: \(\displaystyle \sum_{n=2}^{\infty}ln\left(1 - \frac{1}{n^2}\right) = -ln(2)\)

Proof: If we expand the sum out term-wise, we get:

\(\displaystyle \sum_{n=2}^{\infty}ln\left(1 - \frac{1}{n^2}\right) = ln\left(1 - \frac{1}{4}\right) + ln\left(1 - \frac{1}{9}\right) + \cdots\).

Using the property of the natural logarithm that \(\displaystyle ln(a) + ln(b) = ln(ab)\) we get that:

\(\displaystyle \sum_{n=2}^{\infty}ln\left(1 - \frac{1}{n^2}\right) = ln\left(\prod_{n=2}^{\infty}\left(1 - \frac{1}{n^2}\right)\right)\)

Now we need to find the infinite product, \(\displaystyle \prod_{n=2}^{\infty}\left(1 - \frac{1}{n^2}\right)\).

Claim: Let \(\displaystyle k\) be an integer such that \(\displaystyle k\geq 2\). Then the partial product is given by \(\displaystyle \prod_{n=2}^{k}\left(1 - \frac{1}{n^2}\right) = \frac{k+1}{2k}\)

We can prove this by induction. When \(\displaystyle k=2\), we get that:

\(\displaystyle \prod_{n=2}^{2}\left(1 - \frac{1}{n^2}\right) = \frac{3}{4} = \frac{2 + 1}{2*2}\).

Now, suppose that \(\displaystyle \prod_{n=2}^{k}\left(1 - \frac{1}{n^2}\right)\) for some integer k greater than 2, then:

\(\displaystyle \prod_{n=2}^{k+1}\left(1 - \frac{1}{n^2}\right) = \prod_{n=2}^{k}\left(1 - \frac{1}{n^2}\right)*\left(1 - \frac{1}{(k+1)^2}\right)\)

By the inductive hypothesis, we get:

\(\displaystyle \prod_{n=2}^{k+1}\left(1 - \frac{1}{n^2}\right) = \frac{k + 1}{2k}*\left(1 - \frac{1}{(k+1)^2}\right) = \frac{k+1}{2k} - \frac{1}{2k(k+1)} = \frac{(k+1)^2 - 1}{2k(k+1)} = \frac{k^2 + 2k}{2k(k+1)} = \frac{(k+1) + 1}{2(k+1)}\)

and our claim is proved.

If we let \(\displaystyle k\) go to infinity, we get \(\displaystyle \lim_{k\to \infty}\frac{k + 1}{2k} = \frac{1}{2}\).

Finally,

\(\displaystyle \sum_{n=2}^{\infty}ln\left(1 - \frac{1}{n^2}\right) = ln\left(\prod_{n=2}^{\infty}\left(1 - \frac{1}{n^2}\right)\right) = ln\left(\frac{1}{2}\right) = ln(1) - ln(2) = -ln(2)\)

and we are done.[/sp]