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

## Homework Statement

Σ (-1)^n(ln(n)/root(n))

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.

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SammyS
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## Homework Statement

Σ (-1)^n(ln(n)/root(n))

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.

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

Dick
Homework Helper
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.

Last edited:
SammyS
Staff Emeritus
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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
Homework Helper
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
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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→∞ an = 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 an 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