The sum $\frac{1}{\ln(2)}+\frac{1}{\ln(3)}+\frac{1}{\ln(4)}+...$ is divergent. Find the regularized evaluation

Question

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

$$1+2+3+...=-\frac{1}{12}$$

Answer 1

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 $\displaystyle \lim_{N\to+\infty}\int_0^N \zeta(x)-1-\frac 1{x-1}\,dx+\ln(N-1)$.

Answer 2

Since $$ \ln k \le k \quad\forall k \ge 2, $$ it follows that $$ \sum_{k=2}^\infty\frac{1}{\ln k}\ge \sum_{k=2}^\infty\frac{1}{k}=\infty. $$

Answer 3

$$\sum^\infty_{n=1}\dfrac{1}{\ln(n+1)}$$ The comparison test, available here, shows that this is divergent.

Answer 4

$\ln(x)$ 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.