tough sum/product series involving reciprocal of ln

I ran across an interesting series. I looked at it and must admit, I do not even know where to begin. I tried playing around with it, but to no avail.

Here it is. Perhaps it isn't doable.

$$\displaystyle \frac{1}{\ln(2)}+\frac{1}{\ln(2)\ln(3)}+\frac{1}{\ln(2)\ln(3)\ln(4)}+\cdots$$

I ran this through Maple. It does appear to converge.

It gave me $4.934269949\dots$

Can this even be evaluated in some clever manner?

-
Looks unlikely to have a closed form to me, with the multiple logarithms and all. –  Ｊ. Ｍ. Dec 3 '11 at 12:13
It's easy to see that it indeed converges (by the Ratio Test, say). –  David Mitra Dec 3 '11 at 12:21

I think the key is "first to generalize". Clearly the infinite product of logs of consecutive arguments is divergent, but as in the problem of summing-of-like powers, which leads to the bernoulli-numbers, it can be meaningful to provisorically assume a constant expression for this. Say we assume a function $\small P(a) = \ln(1+a) \cdot \ln(1+a+1) \cdot \ln(1+a+2) \cdot \ldots$, then we could write the above sum as $$\small S(2) = {P(2) \over P(1)} + {P(3) \over P(1)} + {P(4) \over P(1)} + \ldots$$ Now say $\small f(a) = {P(a) \over P(1)}$ then the other form $$\small S(a) = f(a)+ f(a+1) +f(a+2) + \ldots$$ hints to an expression involving bernoulli-polynomials or along the EulerMaclaurin-summation-formula. It requires then a "transfer"-function $\small f(x+1)=\varphi(f(x))$ and the mechanism of indefinite summation.