3
$\begingroup$

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

$\endgroup$
8
  • $\begingroup$ You mean does it converge, don't you? $\endgroup$ Dec 28 '14 at 7:33
  • $\begingroup$ ^It is possible he wants to if there is a closed form. It seems to converge to roughly $4.93427....$ $\endgroup$
    – JimmyK4542
    Dec 28 '14 at 7:34
  • $\begingroup$ How find this sum without the use of programs? $\endgroup$
    – piteer
    Dec 28 '14 at 7:38
  • $\begingroup$ 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 ? $\endgroup$ Dec 28 '14 at 7:41
  • $\begingroup$ Series is infinite $\endgroup$
    – piteer
    Dec 28 '14 at 7:45
3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Does your last comment imply that a closed form is known? $\endgroup$
    – guest
    Dec 28 '14 at 8:25
  • $\begingroup$ @guest No. I am only stating that should such a closed form exist, obtaining it is very likely to be much more difficult than something as simple as what I wrote for convergence. $\endgroup$
    – heropup
    Dec 28 '14 at 8:27
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.