Does the series $\sum \frac{1}{\ln^3(k)}$ converge or diverge? My question is if $\displaystyle\sum\limits_{k=2}^\infty \frac{1}{\ln^3(k)}$ converges or diverges.
I could so far use the Integral Test, and solved $\displaystyle\int_2^\infty\dfrac{1}{\ln^3(x)}dx$, using Integration by parts in $\displaystyle\int_2^b\dfrac{1}{\ln^3(x)}dx$ by letting $u=\dfrac{1}{\ln^2(x)},~dv=dx$. It diverges and Wolfram also says the series diverges, so I could get the result this way.
However, I feel there's a better approach using another test, and also have the series $\displaystyle\sum\limits_{k=2}^\infty \frac{1}{\ln^s(k)}$, which might be more difficult to integrate. I would be doing that integral over and over again. Knowing the inequalities $\ln(k)<k^p<b^k<k!<k^k$, is there a bound to actually solve any of the series posted above using a direct comparison?
 A: Hint: show that for $x>100$, $\ln(x)<x^{1/3}$ holds.

Another approach: $\forall t>0$ we have $\ln(t)<t$. Now take $t=x^{1/3}$ and get $ \ln(x)<3x^{1/3}$
A: HINT
According to the Cauchy Condensation test, the proposed series converges (diverges) iff the following series converges (diverges):
\begin{align*}
\sum_{k=2}^{\infty}2^{k}a_{2^{k}} = \sum_{k=2}^{\infty}\frac{2^{k}}{\ln^{3}(2^{k})} = \frac{1}{\ln^{3}(2)}\sum_{k=2}^{\infty}\frac{2^{k}}{k^{3}}
\end{align*}
But the last series diverges because its general term does not converge to zero.
Can you justify the last claim?
A: HINT: You know that, for any constant $c$ [such as $c=3$], that the following inequality holds for $k$ large enough:
$$\ln^c(k) < k$$, and thus, for $k$ large enough:
$$\frac{1}{\ln^3(k)} > \frac{1}{k}.$$
So if $\sum_{k\ge2}\frac{1}{k}$ diverges, then so does $\sum_{k \ge 2} \frac{1}{\ln^3(k)}$...
A: Under $\ln(x)\to t$,
$$\int_2^b\dfrac{1}{\ln^3(x)}dx=\int_{\ln(2)}^{\ln(b)}\frac{e^t}{t^3}dt\ge\int_{\ln(2)}^{\ln(b)}\frac{t^2}{t^3}dt=\ln(x)\bigg|_{\ln(2)}^{\ln(b)}=\ln(\ln (b))-\ln(\ln(2))$$
and then letting $b\to\infty$, one has
$$ \int_2^\infty\dfrac{1}{\ln^3(x)}dx=\infty $$
and hence the series diverges.
