Convergence of a series involving logs

Question

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.

Answer 1

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, $$ k\left(1-\frac{\log(k)^{\log(k)}}{\log(k+1)^{\log(k+1)}}\right)\ge2\tag7 $$

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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.

Answer 2

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} $