Showing the following sum is finite

**HTale** · February 1st 2009, 12:12 PM

Hi guys,

I'm having real trouble proving the following

$-\sum^{\infty}_{n=1}\log \left(1 - \frac{x^2}{n^2}\right) < \infty$

Every avenue I take reaches a dead end!

Thank you all very much in advance,

HTale

**Danny** · February 1st 2009, 12:56 PM

Originally Posted by HTale

Hi guys,

I'm having real trouble proving the following

$-\sum^{\infty}_{n=1}\log \left(1 - \frac{x^2}{n^2}\right) < \infty$

Every avenue I take reaches a dead end!

Thank you all very much in advance,

HTale

If you write out the terms

$- \ln \left( 1 - \frac{x^2}{1^2}\right) - \ln \left( 1 - \frac{x^2}{2^2}\right) - \ln \left( 1 - \frac{x^2}{3^2}\right) - \ln \left( 1 - \frac{x^2}{4^2}\right) \cdots$

Then we can combine the log's giving

$- \ln \prod_{i = 1}^\infty \left( 1 - \frac{x^2}{n^2}\right)$

The product has a nice closed form

$\sin \pi x = \pi x \prod_{i = 1}^\infty \left( 1 - \frac{x^2}{n^2}\right)$

so what you have is

$- \ln \frac{\sin \pi x}{\pi x}$ .

It might be easier to deal with now.

**Opalg** · February 1st 2009, 01:00 PM

I think you should assume that |x| < 1 (otherwise the first term of the series will not be defined, being the log of a negative number).

Now use the power series expansion for –log(1–t) (valid when 0<t<1):

$\begin{aligned}-\log(1-t) &= t + \frac{t^2}2 + \frac{t^3}3 + \frac{t^4}4 + \ldots \\<br /> &\leqslant t + t^2 + t^3 + t^4 + \ldots = \frac t{1-t}.\end{aligned}$

Substitute $t = x^2/n^2$ to see that $-\log\Bigl(1-\frac{x^2}{n^2}\Bigr) \leqslant \frac{x^2}{n^2-x^2}$ and use the comparison test with $\sum\frac1{n^2}$ .

**HTale** · February 1st 2009, 03:01 PM

Originally Posted by Opalg

I think you should assume that |x| < 1 (otherwise the first term of the series will not be defined, being the log of a negative number).

Now use the power series expansion for –log(1–t) (valid when 0<t<1):

$\begin{aligned}-\log(1-t) &= t + \frac{t^2}2 + \frac{t^3}3 + \frac{t^4}4 + \ldots \\<br /> &\leqslant t + t^2 + t^3 + t^4 + \ldots = \frac t{1-t}.\end{aligned}$

Substitute $t = x^2/n^2$ to see that $-\log\Bigl(1-\frac{x^2}{n^2}\Bigr) \leqslant \frac{x^2}{n^2-x^2}$ and use the comparison test with $\sum\frac1{n^2}$ .

Thanks for that Opalg. This was incidently one of the routes I took. But, I convinced myself otherwise as my problem here is that the comparision test is set for positive series. So, we have to set $x < n$ (which is what we already have). However, for some fixed $x$ , it's not always the case that

$\frac{x^2}{n^2-x^2} \leq \frac{1}{n^2}$

what do I do/say in my argument to fix this?

Thanks in advance,

HTale

**Opalg** · February 2nd 2009, 12:11 AM

Originally Posted by HTale

Thanks for that Opalg. This was incidentally one of the routes I took. But, I convinced myself otherwise as my problem here is that the comparison test is set for positive series. So, we have to set $x < n$ (which is what we already have). However, for some fixed $x$ , it's not always the case that

$\frac{x^2}{n^2-x^2} \leq \frac{1}{n^2}$

what do I do/say in my argument to fix this?

Convergence (or otherwise) of a series is unaffected by what happens to finitely many terms at the start of the series. So to apply the comparison test, for some fixed value of x, there is no loss in assuming that n>2x. Then $\frac{x^2}{n^2-x^2}$ will be less than some multiple of $\frac{1}{n^2}$ .

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