An infinite series for $\ln 2$ and taking limits of summations Example $1$.$$\tfrac 14\text{Log }2=\sum\limits_{k=1}^\infty\frac 1{\left\{2(2k-1)\right\}^3-2(2k-1)}\tag1$$
Proof. The right side above is equal to $\tfrac 12\{\varphi(2)-\varphi(4)\}$. Hence, the result follows from the two identities$$\text{Log }2=\tfrac 12\varphi(2)$$$$\tfrac 32\text{Log }2=\varphi(4)$$


Question: How do you prove $(1)$?

I started off with $\varphi(2,n)$ and $\varphi(4,n)$; subtracted them, and tried to show that together, the limit was equal to $\tfrac 12\text{Log }2$. However, this is all I have$$\begin{align*}\varphi(2,n)-\varphi(4,n) & =\sum\limits_{k=1}^n\left\{\frac 1{2k-1}+\frac 1{2k+1}-\frac 1k\right\}-\sum\limits_{k=1}^n\left\{\frac 1{4k-1}+\frac 1{4k+1}-\frac 1{2k}\right\}\\ & \\ & =\sum\limits_{k=1}^n\frac 1{2k-1}+\sum\limits_{k=1}^n\frac 1{2k+1}-\sum\limits_{k=1}^n\frac 1k-\sum\limits_{k=1}^n\frac 1{4k-1}-\sum\limits_{k=1}^n\frac 1{4k+1}+\sum\limits_{k=1}^n\frac 1{2k}\\ & \\ & =\sum\limits_{k=1}^n\frac 1{2k-1}+\sum\limits_{k=1}^n\frac 1{2k+1}-\sum\limits_{k=1}^n\frac 1{4k-1}-\sum\limits_{k=1}^n\frac 1{4k+1}-\frac 12\sum\limits_{k=1}^n\frac 1k\end{align*}$$However, it's after this step that I'm not sure what to do. If you split up twice the RHS of $(1)$, you get a $1/(4k-3)$ term, a $1/(4k-1)$ term and a $1/(2k-1)$ term. I'm puzzled on how they got there.

Also, I should probably add the definitions of the notations$$\lim\limits_{n\to\infty}\varphi(a,n)=\varphi(a)$$$$\varphi(a,n)=1+2\sum\limits_{k=1}^n\left\{\frac 1{(ak)^3-ak}\right\}$$Where $a$ is greater than one.
Sorry for the increase in summation questions, I'm analyzing Ramanujan's works on Harmonic Series and Inverse Tangent Functions.

Proof:
identity: $$\frac{1}{x^3 -x} = \frac{1}{2(x-1)} + \frac{1}{2(x+1)} -\frac{1}{x}$$
$$\sum\limits_{k=1}^n\frac 1{\{2(2k-1)\}^3-2(2k-1)}=\frac{1}{2}\sum\limits_{k=1}^n\frac 1{4k-3}+\frac{1}{2}\sum\limits_{k=1}^n\frac 1{4k-1}-\frac{1}{2}\sum\limits_{k=1}^n\frac 1{2k-1}$$
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 $(1/4) \log 2$
