Test for absolute convergence

**coobe** · June 24th 2009, 07:45 AM

hello. i have to test for convergence and discuss if its absolte convergent or not:

$\sum \frac{(-1)^n} {\sqrt {n}}$

I think i should use the Leibniz Rule, but not sure how to apply it here...

**apcalculus** · June 24th 2009, 07:56 AM

Originally Posted by coobe

hello. i have to test for convergence and discuss if its absolte convergent or not:

$\sum \frac{(-1)^n} {\sqrt {n}}$

I think i should use the Leibniz Rule, but not sure how to apply it here...

an integral test on $f(x) = \frac{1}{\sqrt{x}}$ should do it.

**Twig** · June 24th 2009, 08:21 AM

Hi

$\displaystyle\sum_{k=1}^{\infty} |\frac{(-1)^{k}}{\sqrt{k}}| =$ $\displaystyle\sum_{k=1}^{\infty} \frac{1}{\sqrt{k}}$

And on this series you could use the integral test on $h(x) = \frac{1}{\sqrt{x}}$

And this series is divergent because the integral is divergent.

But $\displaystyle\sum_{k=1}^{\infty} \frac{(-1)^{k}}{\sqrt{k}}$ is convergent by Leibniz convergene theorem.

Because the series is alternating, decreasing and $\lim_{n\to \infty} a_{n} = 0$

**matheagle** · June 25th 2009, 10:37 PM

The absoute series is a p-series.
You should know when a p-series converges or diverges.

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