Solve ∞ Series: Is ∞ \sum(-1)n-1(ln(n))p/n (p>0) Convergent or Divergent?

[Dell]

i have reached a possible answer which is not quite like the one my book has, can someone please tell me if I am wrong.

the question asks if the following converges or diverges.

_∞
$$\sum$$(-1)^n-1(ln(n))^p/n (p>0)
ⁿ⁼¹

i learned that if i have a series which changes sign i take the abs value of $$\sum$$An and test that series, if $$\sum$$|An| converges then An definitely converges, if $$\sum$$|An| diverges, i need to check if lim(n->∞)An=0 and if An>A(n+1) (leibnitz)
i integrated the abs value of function in order to find its behaviour, if i get ∞ then the abs series diverges

$$\int$$((ln(x))^p/x)dx
t=ln(x)
dt=dx/x
x=e^t

$$\int$$t^pdt (from 0 to ∞)

=[lim[b->∞]]$$\frac{1}{p+1}$$(t^p+1)|$$^{b}_{0}$$
=[lim[b->∞]]$$\frac{1}{p+1}$$(b^p+1-0^p+1)= ∞
therefore the abs series diverges so i need to check limAn and An+1

lim An = ln^p(n)/n
n->∞

surely this is dependant on p, i need to be equal to 0, therefore i need n to be bigger than ln^p(n),(since we are talking about a limit of n-> ∞)

ln^p(n) < n
p < log_ln(n)(ln(n))

is this correct, or will ln^p(n) < n always be true since n->∞?

now how do i prove than An > An+1

[tex]\frac{ln^p(n+1)}{n+1}[/tex] < [tex]\frac{ln^p(n)}{n}[/tex]
?

alternatively, do you see any better ways to solve this??

 

[lippyka]

Yes, your answer is correct. To prove that An > An+1, you can use the fact that ln(x) is a decreasing function for x > 0 so lnp(n+1) < lnp(n). This implies that An+1 < An and thus the series diverges.