Convergence/Divergence of (-1)^n(ln(n)/root(n))

[Somnus]

Homework Statement

Σ (-1)^n(ln(n)/root(n))

Homework Equations

None

The Attempt at a Solution

Tried all the tests, and the only one that makes sense to use is the Alternating Serioes Test. I know the lmit of ln (n)/ root (n) approaches 0. But not sure how to prove that ln (n) / root n is greater than ln n+1 / root (n+1)

I know the series converges, but not sure of the process in finding so.

 

[SammyS]

Homework Statement

Σ (-1)^n(ln(n)/root(n))

Homework Equations

None

The Attempt at a Solution

Tried all the tests, and the only one that makes sense to use is the Alternating Serioes Test. I know the lmit of ln (n)/ root (n) approaches 0. But not sure how to prove that ln (n) / root n is greater than ln n+1 / root (n+1)

I know the series converges, but not sure of the process in finding so.

Hello Somnus. Welcome to PF !

I'm pretty sure that the series does not converge.

 

[Somnus]

According to the answers in my book, the series converges.

 

[Dick]

According to the answers in my book, the series converges.

To apply the alternating series test you could show that for x large enough that the derivative of ln(x)/sqrt(x) is negative, so it's a decreasing function. It does converge.

 

[SammyS]

To apply the alternating series test you could show that for x large enough that the derivative of ln(x)/sqrt(x) is negative, so it's a decreasing function. It does converge.

OK ! I stand corrected.

I'll take Dick's word over that of WolframAlpha. I had based my earlier comment on a result from WolframAlpha.

The convergence must be awfully "slow".

 

[Dick]

OK ! I stand corrected.

I'll take Dick's word over that of WolframAlpha. I had based my earlier comment on a result from WolframAlpha.

The convergence must be awfully "slow".

It's not absolutely convergent if that's what you mean by slow. WA is easy to confuse if you don't think about the question you send it.

 

[SammyS]

It's not absolutely convergent if that's what you mean by slow. WA is easy to confuse if you don't think about the question you send it.

No.

I know it's not absolutely convergent.

I just gave it the sum. http://www2.wolframalpha.com/input/?i=%CE%A3+%28-1%29^n%28ln%28n%29%2Fsqrt%28n%29%29+

WA also gives a partial sum which I took out to 1200 terms.

 

[Deveno]

Wolfram|Alpha appears to be just wrong on this one (and it's not a matter of "bad input").

provided one has shown that lim n→∞ a_n = 0 (and L'Hopital's Rule works well here), than Liebnitz's test applies, and to show monoticity, one can examine the derivative of:

a(x) = ln(x)/√x.

it's not hard to show that there is an N such that a_n is montonically decreasing for all n > N (N is even fairly small).

one caveat: the starting integer is not explicitly given, although given the nature of the series, 2 seems a reasonable assumption.

 

[Dick]

I have seen WA announce that a divergent series is convergent, giving a finite sum by reordering the terms. Perhaps here it is reordering the terms to turn a convergent series into a divergent one. I think it plays a little loosely with conditionally convergent series.

 

[Deveno]

I have seen WA announce that a divergent series is convergent, giving a finite sum by reordering the terms. Perhaps here it is reordering the terms to turn a convergent series into a divergent one. I think it plays a little loosely with conditionally convergent series.

i think it's just a bad algorithm somewhere (it's probably comparing it to some other divergent series, and mis-matching the terms. for conditionally convergent series, the way you sum the series matters very much). the fact that the first non-zero term is a₂ may have something to do with this. i sent WA an e-mail, they may investigate, they may not.