For What Value of P is $\sum (-1)^{n-1}\frac{(\ln(n))^p}{n}$ convergent? The Question 
for what value of p is $\sum (-1)^{n-1}\frac{(\ln(n))^p}{n}$ convergent?
My Work
It's an alternating series, so I have to show which p values it's limit to infinity go to $0$ and I have to show for what values of p it is decreasing.
Showing which Values it is decreasing on
I first worked out the derivative $\ln(n)^{p-1}(p - \ln(n))$ then I think I showed that for all $p<\ln(n)$ it is less than $0$.
Showing where the limit is 0
This appears to be true for all values of $p>0$. Using L'hospital's rule we can constantly reduce this problem to some prouduct of $(\ln(n)^{p-something}*p+something)/n$. Eventually $p-something = 1$ and L'hospital's rule will evaluate the limit to zero. Tests on wolframalpha support this.
My Problem
I'm not very sure I did this question correctly. The part about showing value it is decreasing on, I think my book would want a more exact answer. I'm not really sure how to interpret $p<\ln(n)$. I also don't know if my explanation for the limit going to $0$ was correct and if it was I would like to know how to explain it more mathematically.

$f(x) = \dfrac{\ln^p x}{x} \to f'(x) = \dfrac{p(\ln x)^{p-1} - \ln^p x}{x^2} = \dfrac{(\ln x)^{p-1}}{x^2}\cdot (p-\ln x) < 0 \iff p < \ln x \iff e^p < x$. This shows that starting at some index $n > e^p$, the series converges conditionally for all $p$'s.
