<s>[INST] 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.
 [/INST] By partial fraction decomposition
$$ \frac{1}{8x^3-2x} = \frac{1}{(2x-1)2x(2x+1)} = \frac{1}{2}\cdot\frac{1}{2x-1}-\frac{1}{2x}+\frac{1}{2}\cdot\frac{1}{2x+1}\tag{1} $$
hence by replacing $x$ with $2k-1$ we get:
$$ \frac{1}{8(2k-1)^3-2(2k-1)}=\frac{1}{2}\cdot\frac{1}{4k-3}-\frac{1}{4k-2}+\frac{1}{2}\cdot\frac{1}{4k-1}\tag{2} $$
or:
$$ \frac{1}{8(2k-1)^3-2(2k-1)}=\frac{1}{2}\int_{0}^{1}\left(x^{4k-4}-2x^{4k-3}+x^{4k-2}\right)\,dx\tag{3} $$
and by summing both sides of $(3)$ over $k\geq 1$:
$$ \sum_{k\geq 1}\frac{1}{8(2k-1)^3-2(2k-1)}=\frac{1}{2}\int_{0}^{1}\frac{(1-x)^2}{1-x^4}\,dx = \color{red}{\frac{1}{4}\log 2} \tag{4}$$
always by partial fraction decomposition, or through:
$$ \int_{0}^{1}\frac{(1-x)}{(1+x)(1+x^2)}\,dx\stackrel{x\mapsto\tan\theta}{=} \int_{0}^{\pi/4}\frac{\cos\theta-\sin\theta}{\sin\theta+\cos\theta}\,d\theta = \left[\log(\sin\theta+\cos\theta)\right]_{0}^{\pi/4}.\tag{5}$$
</s>