Question

More series issues!

Answer 1

It can't be that $p$ makes the series converge, because this would introduce a zero in the denominator when $n=1$. For a similar reason, $p=0$ would involve an indeterminate term of $0^0$.

That leaves checking what happens when $p>0$. First, consider the function

$f(x)=\dfrac{(\ln x)^p}x$

and its derivative

$f'(x)=\dfrac{p(\ln x)^{p-1}-(\ln x)^p}{x^2}=\dfrac{(\ln x)^{p-1}}{x^2}(p-\ln x)$

$f(x)$ has critical points at $x=1$ and $x=e^p$. (These never coincide because we're assuming $p>0$, so it's always the case that $e^p>1$.) Between these two points, say at $c=\dfrac{e^p}2$, you have $f'(c)=\dfrac{4\ln2}{e^{2p}}(\ln2)^{p-1}$, which is positive regardless of the value of $p$. This means $f(x)$ is increasing on the interval $(1,e^p)$.

Meanwhile, if $x>e^p$ - and let's take $c=2e^p$ as an example - we have $f'(c)=\dfrac{(\ln2+p)^{p-1}}{4e^{2p}}(-\ln2)^{p-1}$, which is negative for all $p>0$. This means $f(x)$ is decreasing for all $x>e^p$, which shows that $\dfrac{(\ln n)^p}n$ is a decreasing sequence for all $n>N$, where $N$ is any sufficiently large number that depends on $p$.

Now, it's also the case that for $p>0$ (and in fact all $p\in\mathbb R$),

$\displaystyle\lim_{n\to\infty}\dfrac{(\ln n)^p}n=0$

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 $p>0$.

For the second question, recall that

$h_n=\displaystyle\sum_{k=1}^n\frac1k=1+\frac12+\cdots+\frac1{n-1}+\frac1n$
$s_n=\displaystyle\sum_{k=1}^n\frac{(-1)^{k-1}}k=1-\frac12+\cdots-\frac1{n-1}+\frac1n$

(note that the above is true for even $n$ only - it wouldn't be too difficult to change things around if $n$ is odd)

It follows that

$h_{2n}=\displaystyle\sum_{k=1}^{2n}\frac{(-1)^{k-1}}k=1+\frac12+\cdots+\frac1{2n-1}+\frac1{2n}$
$s_{2n}=\displaystyle\sum_{k=1}^{2n}\frac{(-1)^{k-1}}k=1-\frac12+\cdots+\frac1{2n-1}-\frac1{2n}$

Subtracting $h_{2n}$ from $s_{2n}$, you have

$\displaystyle s_{2n}-h_{2n}=(1-1)+\left(-\frac12-\frac12\right)+\left(\frac13-\frac13\right)+\left(-\frac14-\frac14\right)+\cdots+\left(\frac1{2n-1}-\frac1{2n-1}\right)+\left(-\frac1{2n}-\frac1{2n}\right)$
$s_{2n}-h_{2n}=-1-\dfrac12-\cdots-\dfrac2{2n}$
$s_{2n}-h_{2n}=-\left(1+\dfrac12+\cdots+\dfrac1n\right)$
$s_{2n}-h_{2n}=-h_n$
$\implies s_{2n}=h_{2n}-h_n$

as required. Notice that assuming $n$ is odd doesn't change the result; the last term in $h_{2n}$ ends up canceling with the corresponding term in $s_{2n}$ regardless of the parity of $n$.