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.


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: $$ \int_{0}^{1}\frac{(1-x)}{(1+x)(1+x^2)}\,dx\stackrel{x\mapsto\tan\theta}{=} \int_{0}^{\pi/4}\frac{\cos\theta-\sin\theta}{\sin\theta+\cos\theta}\,d\theta = \left[\log(\sin\theta+\cos\theta)\right]_{0}^{\pi/4}.\tag{5}$$

  • $\begingroup$ Is there a way to prove this without using integrals? I haven't quite mastered them yet. I'm sticking with manipulating summing series for now. $\endgroup$ – Crescendo Jun 23 '17 at 13:13
  • $\begingroup$ @Crescendo: if you are able to show that $H_n=\sum_{k=1}^{n}\frac{1}{k}=\log(n)+O(1)$ the other answer provides a good way without integrals. The asymptotic approximation for the harmonic number $H_n$ can be derived from creative telescoping, for instance. $\endgroup$ – Jack D'Aurizio Jun 23 '17 at 13:16

Let $S(n)=\sum_{k=1}^n\frac {1}{k}.$ We have $S(n)=\gamma +\log n +O(n^{-1})$ as $n\to \infty,$ where $\gamma$ is Euler's constant.

Let $T(n)= \sum_{j=1}^n\frac {1}{2j-1}.$ We have $T(n)=S(2n)-\frac {1}{2}S(n)=\frac {1}{2} \gamma +\log 2+\frac {1}{2}\log n+O(n^{-1}).$

We have $$\sum_{k=1}^n\frac {1}{ [2(2k-1)]^3-2(2k-1)}=\frac {1}{2}\sum_{k=1}^n \frac {1}{4k-3}+\frac {1}{4k-1}-\frac {2}{4k-2}=$$ $$=\frac {1}{2}(\;T(2n)-T(n)\;)=$$ $$=\frac {1}{2}\left(\frac {1}{2}\log 2n-\frac {1}{2}\log n\right)+O(n^{-1})=\frac {1}{4}\log 2+O(n^{-1}).$$


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 $(1/4) \log 2$


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