It can't be that [tex]p<0[/tex] makes the series converge, because this would introduce a zero in the denominator when [tex]n=1[/tex]. For a similar reason, [tex]p=0[/tex] would involve an indeterminate term of [tex]0^0[/tex].
That leaves checking what happens when [tex]p>0[/tex]. First, consider the function
[tex]f(x)=\dfrac{(\ln x)^p}x[/tex]
and its derivative
[tex]f'(x)=\dfrac{p(\ln x)^{p-1}-(\ln x)^p}{x^2}=\dfrac{(\ln x)^{p-1}}{x^2}(p-\ln x)[/tex]
[tex]f(x)[/tex] has critical points at [tex]x=1[/tex] and [tex]x=e^p[/tex]. (These never coincide because we're assuming [tex]p>0[/tex], so it's always the case that [tex]e^p>1[/tex].) Between these two points, say at [tex]c=\dfrac{e^p}2[/tex], you have [tex]f'(c)=\dfrac{4\ln2}{e^{2p}}(\ln2)^{p-1}[/tex], which is positive regardless of the value of [tex]p[/tex]. This means [tex]f(x)[/tex] is increasing on the interval [tex](1,e^p)[/tex].
Meanwhile, if [tex]x>e^p[/tex] - and let's take [tex]c=2e^p[/tex] as an example - we have [tex]f'(c)=\dfrac{(\ln2+p)^{p-1}}{4e^{2p}}(-\ln2)^{p-1}[/tex], which is negative for all [tex]p>0[/tex]. This means [tex]f(x)[/tex] is decreasing for all [tex]x>e^p[/tex], which shows that [tex]\dfrac{(\ln n)^p}n[/tex] is a decreasing sequence for all [tex]n>N[/tex], where [tex]N[/tex] is any sufficiently large number that depends on [tex]p[/tex].
Now, it's also the case that for [tex]p>0[/tex] (and in fact all [tex]p\in\mathbb R[/tex]),
[tex]\displaystyle\lim_{n\to\infty}\dfrac{(\ln n)^p}n=0[/tex]
So you have a series of a sequence that in absolute value is decreasing and converging to 0. The alternating series then says that the series must converge for all [tex]p>0[/tex].
For the second question, recall that
[tex]h_n=\displaystyle\sum_{k=1}^n\frac1k=1+\frac12+\cdots+\frac1{n-1}+\frac1n[/tex]
[tex]s_n=\displaystyle\sum_{k=1}^n\frac{(-1)^{k-1}}k=1-\frac12+\cdots-\frac1{n-1}+\frac1n[/tex]
(note that the above is true for even [tex]n[/tex] only - it wouldn't be too difficult to change things around if [tex]n[/tex] is odd)
It follows that
[tex]h_{2n}=\displaystyle\sum_{k=1}^{2n}\frac{(-1)^{k-1}}k=1+\frac12+\cdots+\frac1{2n-1}+\frac1{2n}[/tex]
[tex]s_{2n}=\displaystyle\sum_{k=1}^{2n}\frac{(-1)^{k-1}}k=1-\frac12+\cdots+\frac1{2n-1}-\frac1{2n}[/tex]
Subtracting [tex]h_{2n}[/tex] from [tex]s_{2n}[/tex], you have
[tex]\displaystyle s_{2n}-h_{2n}=(1-1)+\left(-\frac12-\frac12\right)+\left(\frac13-\frac13\right)+\left(-\frac14-\frac14\right)+\cdots+\left(\frac1{2n-1}-\frac1{2n-1}\right)+\left(-\frac1{2n}-\frac1{2n}\right)[/tex]
[tex]s_{2n}-h_{2n}=-1-\dfrac12-\cdots-\dfrac2{2n}[/tex]
[tex]s_{2n}-h_{2n}=-\left(1+\dfrac12+\cdots+\dfrac1n\right)[/tex]
[tex]s_{2n}-h_{2n}=-h_n[/tex]
[tex]\implies s_{2n}=h_{2n}-h_n[/tex]
as required. Notice that assuming [tex]n[/tex] is odd doesn't change the result; the last term in [tex]h_{2n}[/tex] ends up canceling with the corresponding term in [tex]s_{2n}[/tex] regardless of the parity of [tex]n[/tex].
