We know Ramanujan's $$\phi(a,n) = 1+ 2\sum_{k =0}^{n}{\frac{1}{(ak)^3-ak}}$$
How can I prove
$$\sum_{k=1}^n\frac{1}{n+k} = \frac{n}{2n+1} + \sum_{k=1}^n(\frac{1}{(2k)^3-2k})$$
Don't know how to prove using above $\phi(2, \infty) = \log(4) $
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2What is the summand of your last expression ? – Claude Leibovici Sep 28 '21 at 09:59
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Welcome MSE You had to add how $\phi(a,n)$ is $\phi(a)$ when $n\to\infty$ so that people can understand what is $\phi(2)$ – Sep 28 '21 at 11:03
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$$\phi(a, n) = 1+ 2\sum_{k=1}^n\frac{1}{(ak)^3-ak}$$
Let's denote $$\phi(a) = \lim_{n\to\infty}\phi(a,n)$$
Now coming back to the series to prove:
$$\color{green}{\sum_{k=1}^n\frac{1}{n+k} = \frac{n}{2n+1} + \sum_{k=1}^n\left(\frac{1}{(2k)^3-2k}\right)}$$
Consider this partial fraction
$$\begin{align*} \color{blue}{\frac{1}{p^3-p} =\frac{1}{2(p-1)} +\frac{1}{2(p+1)}-\frac{1}{p}} \end{align*}$$
Let: $y =2k$
$$\begin{align*} S(k,n) & = \frac{1}{2}\sum_{k=1}^n\frac{1}{2k-1} + \frac{1}{2}\sum_{k=1}^n\frac{1}{2k+1}-\frac{1}{2}\sum_{k=1}\frac{1}{k} + \frac{n}{2n+1}\\ & = \sum_{k=1}^n\frac{1}{2k-1}-\frac{1}{2}\sum_{k=1}^n\frac{1}{k}\\ & = \sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k}\\ & = \sum_{k=1}^{n}\frac{1}{n+k} \end{align*}$$
Proved.
Now, I believe you can prove: $$\begin{align*} \frac{1}{2}\phi(2, n\to\infty)\\ = \frac{1}{2}\phi(2)\\ = \ln(2)\\ & \end{align*}$$
$2\ln(2) = \ln(4) = \phi(2)$
It is better to write for $\ln(2) = \frac{1}{2}\phi(2)$
because in Number-Theory $\log2$ is one of the fascinating things.