# Find $\frac{1}{\log 2}+\frac{1}{(\log 2)(\log 3)}+\frac{1}{(\log 2)(\log 3)(\log 4)}+ \cdots$

Is it possible to calculate the sum of $\dfrac{1}{\log 2}+\dfrac{1}{(\log 2)(\log 3)}+\dfrac{1}{(\log 2)(\log 3)(\log 4)}+ \cdots$?

• You mean does it converge, don't you? Dec 28, 2014 at 7:33
• ^It is possible he wants to if there is a closed form. It seems to converge to roughly $4.93427....$ Dec 28, 2014 at 7:34
• How find this sum without the use of programs? Dec 28, 2014 at 7:38
• Computing it shows a very fast convergence. Using $15$ terms already gives $4.93419$; $20$ terms lead to the value given by JimmyK4542. Now, is there a closed form ? Dec 28, 2014 at 7:41
• Series is infinite Dec 28, 2014 at 7:45

Convergence is easily proven since the series can be written as \begin{align*} S &= \sum_{k=2}^\infty \biggl( \prod_{j=2}^k \log j \biggr)^{-1} \\ &= (\log 2)^{-1} \left( 1 + \sum_{k=3}^\infty \biggl( \prod_{j=3}^k \log j \biggr)^{-1} \right) \\ &< (\log 2)^{-1} \left( 1 + \sum_{k=1}^\infty (\log 3)^{-k}\right) \\ &= \frac{\log 3}{\log 2(\log 3 - 1)}.\end{align*} A closed form value is substantially more difficult to obtain but I will leave this as an establishment of convergence.
Partial answer: let $$f(x)=\dfrac{1}{\log x}+\dfrac{1}{(\log x)(\log (x+1))}+\dfrac{1}{(\log x)(\log (x+1))(\log (x+2))}+ \cdots$$ then \begin{align} f(x)&=\dfrac{1}{\log x}+\dfrac{1}{(\log x)(\log (x+1))}+\dfrac{1}{(\log x)(\log (x+1))(\log (x+2))}+ \cdots\\ f(x)&=\dfrac{1}{\log x}\left(1+\dfrac{1}{(\log (x+1))}+\dfrac{1}{(\log (x+1))(\log (x+2))}+ \cdots\right)\\ f(x)&= \frac{1+f(x+1)}{\log(x)}\\ \implies f(x+1)&=\log(x) f(x)-1 \end{align} So we now have a recurrence relation for the function, and we are seeking $f(2)$. By repeatedly reducing $f(x+n+1)$, we also obtain $$f(x+n)=\left(\prod_{k=0}^{n-1}\log(x+k)\right) f(x)-\sum_{k=0}^{n-1}\prod_{i=1}^{n-1}\log(x+i)$$
So if we can come up with $f(x)$ for some integer $x$, we can come up with $f(x)$ for any integer $x$. I can't seem to come up with a value for anything though, so hopefully someone else can.