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 [math]\sum_{n=2}^{\infty}ln\left(1 - \frac{1}{n^2}\right)[/math].

Claim: [math]\sum_{n=2}^{\infty}ln\left(1 - \frac{1}{n^2}\right) = -ln(2)[/math]

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

[math]\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[/math].

Using the property of the natural logarithm that [math]ln(a) + ln(b) = ln(ab)[/math] we get that:

[math]\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)[/math]

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

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

We can prove this by induction. When [math]k=2[/math], we get that:

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

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

[math]\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)[/math]

By the inductive hypothesis, we get:

[math]\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)}[/math]

and our claim is proved.

If we let [math]k[/math] go to infinity, we get [math]\lim_{k\to \infty}\frac{k + 1}{2k} = \frac{1}{2}[/math].

Finally,

[math]\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)[/math]

and we are done.[/sp]