4
$\begingroup$

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?

  • 0
    It's easy to see that it indeed converges (by the Ratio Test, say).2011-12-03

1 Answers 1

1

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.

Perhaps this hint is already enough; I think I'll try some more steps myself later today and extend this answer if needed.

  • 0
    Thanks for the input. I had a feeling it may not have a closed form. I figured if anyone could solve it, though, it would be someone on this site :) Cheers2011-12-04