Convergence of a series involving logs 
Show that $\displaystyle\sum_{k=2}^\infty \dfrac{1}{(\log k)^{\log k}}$ converges using Raabe's test.

We want to evaluate $\lim\limits_{k\to\infty} k(1-\dfrac{\frac{1}{(\log (k+1))^{\log (k+1)}}}{\frac{1}{(\log k)^{\log k}}}) = \lim\limits_{k\to\infty} k(1-\dfrac{(\log k)^{\log k}}{(\log (k+1))^{\log (k+1)}}) = \lim\limits_{k\to\infty} k(\dfrac{(\log (k+1))^{\log (k+1)}-(\log k)^{\log k}}{(\log (k+1))^{\log (k+1)}}).$
I need to show that the above limit is greater than $1$, or, if it is equal to $1$, then $\sup_{k\in\mathbb{N}}k|k(1-\dfrac{a_{k+1}}{a_k})-1| < \infty, a_k := \dfrac{1}{(\log k)^{\log k}}$
I know that $\dfrac{1}{(\log k)^{\log k}} = e^{-\log k\log (\log k)},$ but I'm not sure how this can be useful. 
 A: If One MUST Use Raabe's Test
For $k\ge3$,
$$
\begin{align}
k\left(1-\frac{\log(k)^{\log(k)}}{\log(k+1)^{\log(k+1)}}\right)
&\ge k\left(1-\frac{\log(k)^{\log(k)}}{\log(k)^{\log(k+1)}}\right)\tag1\\[6pt]
&=k\left(1-\log(k)^{\log(k)-\log(k+1)}\right)\tag2\\[9pt]
&\ge k\left(1-\log(k)^{-\frac1{k+1}}\right)\tag3\\[6pt]
&=k\left(1-e^{-\frac{\log(\log(k))}{k+1}}\right)\tag4\\
&\ge k\left[1-\frac1{1+\frac{\log(\log(k))}{k+1}}\right]\tag5\\
&=\frac{k}{k+1}\frac{\log(\log(k))}{1+\frac{\log(\log(k))}{k+1}}\tag6
\end{align}
$$
Explanation:
$(1)$: $\log(k+1)\gt\log(k)$
$(2)$: $a^b/a^c=a^{b-c}$
$(3)$: $\frac1{k+1}\le\log(k+1)-\log(k)$
$(4)$: $\log(k)=e^{\log(\log(k))}$
$(5)$: $e^{-x}\le\frac1{1+x}$
$(6)$: algebra
Both $\frac{k}{k+1}$ and $\frac1{1+\frac{\log(\log(k))}{k+1}}$ increase to $1$, while $\log(\log(k))$ increases to $\infty$. Thus, from some point on,
$$
k\left(1-\frac{\log(k)^{\log(k)}}{\log(k+1)^{\log(k+1)}}\right)\ge2\tag7
$$

Simpler Approach
As I mentioned in a comment, when $\log(k)\ge e^2$, we have
$$
\begin{align}
\frac1{\log(k)^{\log(k)}}
&\le\frac1{e^{2\log(k)}}\\
&=\frac1{k^2}\tag8
\end{align}
$$
and so we can use the comparison test.
A: Much too messy,
though I might use
$\log(k+1)
=\log(k)+\log(1+1/k)
\lt \log(k)+1/k
$
in a number of places.
Instead,
$ \dfrac{1}{(\log k)^{\log k}}
= \dfrac{1}{e^{\log\log k\log k}}
= \dfrac{1}{k^{\log\log k}}
$
and the sum of these converges
since
$\log\log k
\gt 2
$
for $k > e^{e^2}
\approx 1618
$.
Actually,
all that is needed is
$\log\log k > 1$.
(Added in response to comment.)
This might help.
If $a \to 0$ and
$ab \to 0$
(these are functions of $k$)
then
$(1+a)^b
=e^{b\ln(1+a)}
\approx e^{ab}
\approx 1+ab
$
so
$\begin{array}\\
(\ln(k+1))^{\ln(k+1)}
&\lt(\ln(k)+1/k)^{\ln(k)+1/k}\\
&=(\ln(k))^{\ln(k)+1/k}(1+1/(k\ln(k))^{\ln(k)+1/k}\\
&=(\ln(k))^{\ln(k)}(\ln(k))^{1/k}(1+1/(k\ln(k))^{\ln(k)+1/k}\\
&=(\ln(k))^{\ln(k)}e^{\ln\ln(k)/k}(1+O(1/k))\\
&=(\ln(k))^{\ln(k)}(1+O(\ln\ln(k)/k))\\
\text{so}\\
\dfrac{(\ln(k+1))^{\ln(k+1)}}{(\ln(k))^{\ln(k)}}
&=(1+O(\ln\ln(k)/k))\\
\end{array}
$
