Infinite Series $\sum^{\infty}_{k=2}\frac{1}{k^2\ln(k)}$, Every Test Fails! So we have this infinite series
$$\sum^{\infty}_{k=2}\frac{1}{k^2\ln(k)}$$
It is apparent that $\ln(k)<k$ for $k>2$
Therefore we can compare
$$\sum^{\infty}_{k=2}\frac{1}{k^2\ln(k)} \text{ to } \sum^{\infty}_{k=2}\frac{1}{k^2(k)}=\sum^{\infty}_{k=2}\frac{1}{k^3}$$ 
Which is a convergent $p$ series where $p>1$
Establishing comparison:
$$x^2\ln(x)<x^3,\:\frac{1}{x^2\ln(x)}>\frac{1}{x^3}$$
Therefore comparison test fails $a_n>b_n$ where $b_n$ converges
Limit comparison test:
$$\lim_{k\to\infty}\frac{a_n}{b_n}=\lim_{k\to\infty}\frac{\frac{1}{k^2\ln(K)}}{\frac{1}{k^3}}=\lim_{k\to\infty}\frac{k}{\ln(k)}$$
We must use L'Hopitals Rule, let $f(x)=\frac{x}{\ln(x)}$
$$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{x}{\ln(x)} = \lim_{x\to\infty} \frac{1}{1/x} \to\infty$$
$b_n$ is convergent, so the limit comparison test fails. 
According to Wolfram, the ratio test fails, and $a_n$ cannot be analytically integrated (as far as I can tell). However, Wolfram states that the series IS convergent. 
What test can I use? Or a trick to prove convergence would be appreciated.
Thanks Stax!

One may write, for $k=3,4,\ldots$,
$$
\frac{1}{k^2\ln(k)}<\frac{1}{k^2}
$$ and conclude.
