The sum $\frac{1}{\ln(2)}+\frac{1}{\ln(3)}+\frac{1}{\ln(4)}+...$ is divergent. Find the regularized evaluation By considering the integral Zeta function 
$$F(s)=s+\frac{1}{2^s\ln(2)}+\frac{1}{3^s\ln(3)}+\frac{1}{4^s\ln(4)}+...$$
Evaluate
$$\frac{1}{\ln(2)}+\frac{1}{\ln(3)}+\frac{1}{\ln(4)}+...$$
EDIT:
There has clearly been much confusion here. I am asking for the analytic continuation of the integral Zeta function at 0. I am asking for the sum of the series in the sense that 
$$1+2+3+...=-\frac{1}{12}$$
 A: $\ln(x)$ is smaller than $x$ for all positive values.
By that logic we can say that $\frac{1}{\ln(x)}$ is larger than $\frac{1}{x}$ for all positive x's
This means that $\frac{1}{\ln(2)}+\frac{1}{\ln(3)}+\frac{1}{\ln(4)}+...$ is larger than  $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...$
We know that $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...$ diverges and equals an infinite sum, and since $\frac{1}{\ln(2)}+\frac{1}{\ln(3)}+\frac{1}{\ln(4)}+...$ is larger than that infinite sum, we can conclude that $\frac{1}{\ln(2)}+\frac{1}{\ln(3)}+\frac{1}{\ln(4)}+...$ is an infinite sum.
A: This sum is of course divergent (as proved by others).
But the idea is probably to use zeta regularization and get a formal derivation  like :
\begin{align}
\int_0^{\infty}(\zeta(x)-1)\,dx&=\int_0^{\infty} \sum_{n=2}^\infty \frac 1{n^x}\,dx\\
&=\sum_{n=2}^\infty \int_0^{\infty} e^{-x\ln(n)}\,dx\\
&=-\sum_{n=2}^\infty \left.\frac1{\ln(n)\,n^x}\right|_{x=0}^{\infty}\\
&=\sum_{n=2}^\infty \frac1{\ln(n)}\\
\end{align}
From the simple pole of $\zeta(x)$ at $x=1$ we should write our 'regularized sum' as the Cauchy principal value of the integral :
$$\sum_{n=2}^\infty \frac1{\ln(n)}=PV \int_0^{\infty}(\zeta(x)-1)\,dx\\=-0.243238342890980755415059\cdots$$
(if I didn't make an error...)
For numerical evaluation use $\displaystyle \lim_{N\to+\infty}\int_0^N \zeta(x)-1-\frac  1{x-1}\,dx+\ln(N-1)$.
A: $$\sum^\infty_{n=1}\dfrac{1}{\ln(n+1)}$$
The comparison test, available here, shows that this is divergent.
A: Since
$$
\ln k \le k \quad\forall k \ge 2,
$$
it follows that
$$
\sum_{k=2}^\infty\frac{1}{\ln k}\ge \sum_{k=2}^\infty\frac{1}{k}=\infty.
$$
