Alternating Series and P-series convergence

[kylera]

Alternating Series and P-series "convergence"

I couldn't resist trying out a pun. Anyway, onto the question:

Homework Statement

Test the series for convergence/divergence:
$$\sum^{\infty}_{n=1}\frac{-1^{n-1}}{\sqrt{n}}$$

Homework Equations

Alternating Series Test and possibly p-series test...

The Attempt at a Solution

The expression for an alternating series goes as a(n) = (-1)^(n-1) * b(n). Having said that, it's obvious that b(n) is

$$\frac{1}{\sqrt{n}}$$

which can be re-written as $$\frac{1}{n^\frac{1}{2}}$$. But by the rules of the p-series, since 0.5 is obviously lesser than 1, I came to the conclusion that the series diverges. Instead, the answer states that the series does indeed converge. Can anyone help shed some light in this?

 

[rock.freak667]

I think for the p-series test, the numerator must be '1' only.

 

[Dick]

Forget the p-series test. Concentrate on the alternating series test. b(n) is decreasing towards 0. Sorry, I don't get the 'pun'. Am I being thick?

 

[d_leet]

Sorry, I don't get the 'pun'. Am I being thick?

I don't think you are since I don't get it either.

 

[rock.freak667]

Well usually for the p-series test, the only examples with fractions I've ever seen is where the function is in the form

$$\frac{1}{f(n)}$$

 

[Dick]

Well usually for the p-series test, the only examples with fractions I've ever seen is where the function is in the form

$$\frac{1}{f(n)}$$

Sure, in fact, it applies only to series of the form 1/n^p or n^(p). I wasn't saying you were wrong. It's useful here if you want to discuss absolute convergence. It is useless for the given function.

 

[HallsofIvy]

The "p-series" only applies to positive series. That's why Dick says it is useful to discuss absolute convergence. If your series has both positive and negative terms then it may converge "conditionally". It will converge absolutely only if the series of absolute values converges.

By the "alternating series test", the series
$$\sum \frac{(-1)^n}{\sqrt{n}}$$
converges.

By the "p- series test"
$$\sum \left|\frac{(-1)^n}{\sqrt{n}}\right|= \sum \frac{1}{\sqrt{n}}= \sum \frac{1}{n^{1/2}}$$
does NOT converge and so the original series converges conditionally, not absolutely.