# Am i right here?

1. Apr 16, 2009

### Dell

i have reached a possible answer which is not quite like the one my book has, can someone please tell me if im 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=et

$$\int$$tpdt (from 0 to ∞)

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

lim An = lnp(n)/n
n->∞

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

lnp(n) < n
p < logln(n)(ln(n))

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

now how do i prove than An > An+1

$$\frac{lnp(n+1)}{n+1}$$ < $$\frac{lnp(n)}{n}$$
?????

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