a Series of functions : Does it converge? Does it uniformly converge?

**adam63** · June 25th 2010, 08:15 AM

Hello, I have a question:

$\sum_{n=1}^{\infty} \ln (1+\frac{x^2}{n\ln^2(n)}), x \in (-a,a)$

Does it converge? If so, does it uniformly converge?

Thought of turning the ln into two different ln's, but I really can't find a way to prove a convergence of anything here...

Can you please help me?

Thanks

**Also sprach Zarathustra** · June 25th 2010, 08:30 AM

$lny<y$ for $y>3<br />$
$ln(1+\frac{x^2}{nln^2(n)})<1+\frac{x^2}{nln^2(n)}$

The last term is monotonic decreasing so we can use the integral test... hmmm...

still thinking about this...

**chisigma** · June 25th 2010, 12:15 PM

The term $\frac{x^{2}}{n\ \ln^{2} n}$ is a nonsense for n=1 , so that the series probably is...

$\displaystyle \Lambda = \sum_{n=2}^{\infty} \ln (1+\frac{x^{2}}{n\ \ln^{2} n})$ (1)

The convergence of the 'infinite sum' (1) is equivalent to the convergence of the infinite product...

$\displaystyle e^{\Lambda} = \prod _{n=2}^{\infty} (1+\frac{x^{2}}{n\ \ln^{2} n})$ (2)

... and that is guaranted by the convergence of the series...

$\displaystyle \sum_{n=2}^{\infty} \frac{x^{2}}{n\ \ln^{2} n}$ (3)

... which is true for any value of x. The series (3) converges uniformely in any finite interval of the x and the same holds for the series (1)...

Kind regards

$\chi$ $\sigma$

**adam63** · June 26th 2010, 05:21 AM

Beautiful, thank you very much

!

Only one little thing - how do you explain the "... and that is guaranted by the convergence of the series..." (from (2) to (3) )?

**chisigma** · June 26th 2010, 07:04 AM

The convergence of the series...

$\displaystyle \sum_{n=2}^{\infty} \frac{1}{n\ \ln^{2} n}$ (1)

... can be demostrated applying the 'integral test' considering that is...

$\displaystyle \sum_{n=2}^{\infty} \frac{1}{n\ \ln^{2} n} < 2\ \int_{2}^{\infty} \frac{dx}{x\ \ln^{2} x} = 2\ |- \frac{1}{\ln x}|_{2}^{\infty} = \frac{2}{\ln 2}$ (2)

Kind regards

$\chi$ $\sigma$

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