Is the series $\sum_{n=2}^{\infty}(-1)^n\frac{(\ln n)^a}{n}$ converges? 
Is the series $$\sum_{n=2}^{\infty}(-1)^n\frac{(\ln n)^a}{n}$$
  converges, where $a\in \mathbb{R}$?

Now it is evident that we need to use alternating series test. So we need to show that $a_n=\frac{(\ln n)^a}{n}$ monotonically decreasing and $\lim a_n=0$. 
Let $f(x)=\frac{(\ln x)^a}{x}$. Now $f'(x)=\frac{(\ln x)^{a-1}(a-\ln x)}{x^2}<0$ if $x>e^a$. So for $x>e^a$, $f(x)$ is monotonically decreasing.
Now how can I show that the limit $a_n$ is $0$ as $n\to \infty$? Please help me to solve this last piece in the problem. Thank you. 
 A: let $B$ be an upper bound for $a_n$. then
$$
a_{n^2} = \frac{(\log n^2)^a}{n^2} =\bigg(\frac{2^a}n\bigg)\frac{(\log n)^a}n \lt \frac{2^aB}n
$$
which can be made as small as we like by increasing $n$
A: For $\,a\le0\,$:
$$ 0\,\le\,\frac{(\log{n})^a}{n}=\frac{1}{(\log{n})^{|a|}\,n}\,\le\,\frac{1}{n}\,\Rightarrow\,0\,\le\,\lim_{\small n\rightarrow\infty}\frac{(\log{n})^a}{n}\,\le\,\lim_{\small n\rightarrow\infty}\frac{1}{n}=0 $$
For $\,a\gt0\,$:
$$ 0\,\le\,\frac{(\log{n})^a}{n}\,\le\,\frac{(\log{n})^{\lceil{a}\rceil}}{n}\,\Rightarrow\,0\,\le\,\lim_{\small n\rightarrow\infty}\frac{(\log{n})^a}{n}\,\le\,\lim_{\small n\rightarrow\infty}\frac{(\log{n})^{\lceil{a}\rceil}}{n}=0 $$
Where the last limit could be evaluated by applying L'Hopital's rule $\lceil{a}\rceil$ times:
$$ \lim_{\small n\rightarrow\infty}\frac{(\log{n})^{\small\lceil{a}\rceil}}{n} ={\small{\lceil{a}\rceil}} \lim_{\small n\rightarrow\infty}\frac{(\log{n})^{\small\lceil{a}\rceil-1}}{n} = {\small{\lceil{a}\rceil}\left({\lceil{a}\rceil}-1\right)} \lim_{\small n\rightarrow\infty}\frac{(\log{n})^{\small\lceil{a}\rceil-2}}{n} = {\small\cdots} = \color{red}{{\lceil{a}\rceil}! \lim_{\small n\rightarrow\infty}\frac{1}{n} = 0} 
$$
A: You've already proven that the sequence is monotonic, so we can pick any subsequence we want. I choose the subsequence $n = 2^{2^i}$. What happens when we increase $i$ by $1$? We get that $n$ is squared, which in turn implies that the numerator is multiplied by a factor of $2^a$, while the denominator is multiplied by a factor of $n$. Once $n > 2^a$, this means that we all-in-all multiply the whole fraction by a factor that is smaller than $1$, and gets smaller every time.
A: By L'Hopital
$$\ell=\lim_{\infty}\frac{(\ln n)^a}{n}=\lim_{\infty}\frac{a(\frac1n)(\ln n)^{a-1}}{1}=\lim_{\infty}\frac{a}{\ln n}\frac{(\ln n)^a}{n}=\lim_{\infty}\frac{a}{\ln n}\ell=0\times\ell=0$$
