Question: How do I show that

$$\varphi(2,n)-\varphi(4,n)=2\sum\limits_{k=1}^n\frac 1{\{2(2k-1)\}^3-2(2k-1)}$$

Where$$\varphi(2,n)=1+2\sum\limits_{k=1}^n\frac 1{(2k)^3-2k}$$$$\varphi(4,n)=1+2\sum\limits_{k=1}^n\frac 1{(4k)^3-4k}$$

I started with the LHS, and tried to manipulate it to the RHS.$$\begin{align*}\varphi(2,n)-\varphi(4,n) & =\sum\limits_{k=1}^n\frac 1{2k-1}+\sum\limits_{k=1}^n\frac 1{2k+1}-\sum\limits_{k=1}^n\frac 1{4k-1}-\sum\limits_{k=1}^n\frac 1{4k+1}-\sum\limits_{k=1}^n\frac 1{2k}\\ & \\ & =\left(1+\cdots+\frac 1{2n-1}\right)+\left(\frac 13+\cdots+\frac 1{2n+1}\right)-\left(\frac 13+\cdots+\frac 1{4n-1}\right)-\left(\frac 15+\cdots+\frac 1{4n+1}\right)-\sum\limits_{k=1}^n\frac 1{2k}\end{align*}$$

But, that's as far as I got to. I'm not sure what to do nexy to get the summation. Breaking it apart, we get$$2\sum\limits_{k=1}^n\frac 1{\{2(2k-1)\}^3-2(2k-1)}=\sum\limits_{k=1}^n\frac 1{4k-3}+\sum\limits_{k=1}^n\frac 1{4k-1}-\sum\limits_{k=1}^n\frac 1{2k-1}$$ However, I am not aware as to how the $4k-3$ and $4k-1$ arrived.

  • $\begingroup$ Can you check ? ... because thing are so much nicer if the upper limit of the second sum is $2n$ ... $\varphi(2,n)=1+2\sum\limits_{k=1}^{\color{red}{2}n} \frac 1{(2k)^3-2k}$ $\endgroup$ – Donald Splutterwit Jun 23 '17 at 20:23
  • $\begingroup$ @DonaldSplutterwit Yup, definitely $1+2\sum\limits_{k=1}^n\frac 1{1}{(2k)^3-2k}$ $\endgroup$ – Crescendo Jun 23 '17 at 21:06
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    $\begingroup$ I totally agree with Donald Splutterwit's comment. As written, it cannot work since $\phi(2,1)=\frac{4}{3}$, $\phi(4,1)=\frac{31}{30}$ , $\phi(2,1)-\phi(4,1)=\frac{3}{10}$ while the rhs would be $\frac{1}{3}$. If, as said, the summation goes to $2n$ for $\phi(2,n)$ then $\phi(2,1)=\frac{41}{30}$ and this will be OK. $\endgroup$ – Claude Leibovici Jun 24 '17 at 4:53
  • $\begingroup$ @ClaudeLeibovici Shall we try & persaude the OP to let us answer $\varphi(2,\color{red}{2}n)-\varphi(4,n)=2\sum\limits_{k=1}^n\frac 1{\{2(2k-1)\}^3-2(2k-1)}$ instead ? ...or just Refer to page 30 of plouffe.fr/simon/math/Ramanujan's%20Notebooks%20I.pdf ? $\endgroup$ – Donald Splutterwit Jun 24 '17 at 11:41
  • $\begingroup$ @DonaldSplutterwit That's where I got the problem from. I'm trying to prove it, but it just doesn't add up. $\endgroup$ – Crescendo Jun 24 '17 at 18:21

This is the correct version:

$(1/2)(\varphi(2) - \varphi(4)) = \sum\limits_{k=1}^\infty\frac 1{\{2(2k-1)\}^3-2(2k-1)}$

where $\varphi(a) = \lim_{n \to \infty} \varphi(a,n)$

We can show that $\varphi(2) = 2 \log 2$ and $\varphi(4) = (3/2) \log 2$ and the value of the big sum on the right as $(1/4) \log 2$.

Proof: Everything starts with the identity: $$\frac{1}{x^3 -x} = \frac{1}{2(x-1)} + \frac{1}{2(x+1)} -\frac{1}{x}$$

I am going to keep the upper limit of the sum as $n$ for now.

$$\sum\limits_{k=1}^n\frac 1{\{2(2k-1)\}^3-2(2k-1)}=\frac{1}{2}\sum\limits_{k=1}^n\frac 1{4k-3}+\frac{1}{2}\sum\limits_{k=1}^n\frac 1{4k-1}-\frac{1}{2}\sum\limits_{k=1}^n\frac 1{2k-1}$$

Let RHS = $\frac{y}{2}$

The first two terms of $y$ above give the first two terms below and last term of $y$ above gives the last term below.

$$y= \sum\limits_{k=1}^{4n}\frac{1}{k} - \sum\limits_{k=1}^{2n}\frac{1}{2k} -\left(\sum\limits_{k=1}^{2n}\frac{1}{k} - \sum\limits_{k=1}^{n}\frac{1}{2k}\right)$$

$$y= \left(\sum\limits_{k=1}^{4n}\frac{1}{k} - \log 4n\right) +\log 4n - \frac{1}{2}\left(\sum\limits_{k=1}^{2n}\frac{1}{k} -\log2n\right) - (1/2)\log2n -\left(\sum\limits_{k=1}^{2n}\frac{1}{k}-\log 2n\right) - \log 2n + \frac{1}{2}\left(\sum\limits_{k=1}^{n}\frac{1}{k} - \log n\right) + (1/2)\log n$$

As $n \to \infty$, each of the expressions in parentheses reduces to Euler's constant and cancel out. The remaining terms are:

$$ \log 4n - \log 2n - (1/2)\left(\log 2n - \log n\right) = \log 2 - (1/2) \log 2 = (1/2) \log 2$$

Hence original sum is $(1/4) \log 2$

You can do $\varphi(2)$ and $\varphi(4)$ similarly.


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