Convergence Test: Does \sum(-1)^n-1(lnp(n)/n) {p>0} converge or diverge?

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SUMMARY

The convergence of the series \(\sum(-1)^{n-1}\frac{\ln p(n)}{n}\) for \(p > 0\) can be determined using the positive series \(\sum\frac{\ln p(n)}{n}\). If the positive series diverges, the original series diverges as well. The analysis reveals that for \(p > -1\), the positive series diverges, confirming that the original series also diverges. The Leibniz test can be applied to further analyze the behavior of the series.

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Dell
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i need to tell if the following sequence converges or diverges

[tex]\sum[/tex](-1)n-1lnp(n)/n {p>0}
(n from 1 to infinity)

what i think i need to do is take the positive version of this series

[tex]\sum[/tex]lnp(n)/n {p>0}
(n from 1 to infinity)
if the positive series converges than the original must also, if not i can use leibnitz to tell me its behaviour

the problem it the "p", surely {p>0} is not enough information, do i not need to know if p>1 or p<1?

i can integrate [tex]\int[/tex](lnp(x)/x)dx ===> t=lnx dt=dx/x

[tex]\int[/tex]tpdt
=tp+1/(p+1) =...

...lnp+1(n)|[tex]^{infinity}_{1}[/tex]

if P>-1 [tex]\sum[/tex]lnp(n)/n diverges
if P<-1 [tex]\sum[/tex]lnp(n)/n converges
if p=-1 [tex]\sum[/tex]lnp(n)/n diverges

but i am told that {p>0} therefor i am only left with one option
if P>-1 [tex]\sum[/tex]lnp(n)/n diverges

so now i need to check if:
- lim An = 0
- An+1< An
 
Last edited:
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lim An= ln^p(n)/n = 0

so now how do i prove that An > A(n+1)
 
Last edited:

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