# How to prove this identity of summation of ramanujan?

In the first book of ramanujan him find this identity:

Let k , n $\in\mathbb{N}$* and define $A_k=3^k\left(n+\frac{1}{2}\right)-\frac{1}{2}$. the if r is a positive interger

$$\sum_{k=n+1}^{A_r}\frac{1}{k}=r+2\sum_{k=0}^{r-1}\left(r-k\right)\sum_{j=A_{k-1}+1}^{A_k}\frac{1}{\left(3j\right)^3-3j}$$ where $A_{-1}=0$

I want to know a demonstration for that identity, because he uses it to demonstrate his very popular affirmation:

$$\sum_{n=1}^{1000}\frac{1}{n}=7\frac{1}{2} "very\ nearly"$$

First we need the result from entry $2$ of the reference. Lets start with the parial fractions \begin{eqnarray*} \frac{1}{(3j)^3-3j} =\frac{1/2}{3j-1}-\frac{1}{3j}+\frac{1/2}{3j+1}. \end{eqnarray*} So \begin{eqnarray*} 1+2 \sum_{j=1}^{n} \frac{1}{(3j)^3-3j} =1+ \sum_{j=1}^{n} \left( \frac{1}{3j-1}\color{red}{+\frac{1}{3j}}+\frac{1}{3j+1} \color{red}{-\frac{3}{3j}} \right) \\ = \sum_{j=1}^{3n+1} \frac{1}{j} -\sum_{j=1}^{n} \frac{1}{j} = \sum_{j=n+1}^{3n+1} \frac{1}{j} . \end{eqnarray*} Now with $A_k=3^k n + \frac{1}{2} (3^k-1)$ then $A_{k+1}=3A_k+1$, the above result gives \begin{eqnarray*} \sum_{j=A_k +1}^{A_{k+1}=3A_k+1} \frac{1}{j} =1+ 2 \sum_{j=1}^{A_k} \frac{1}{(3j)^3-3j}. \end{eqnarray*} Now sum the above formula from $k=0,1,\cdots, r-1$ and we have \begin{eqnarray*} \sum_{j=n +1}^{A_{r}} \frac{1}{j} =r+ 2 \sum_{k=0}^{r-1} \sum_{j=1}^{A_k} \frac{1}{(3j)^3-3j}. \end{eqnarray*} Note that the inner sums on the RHS are contained within next sum, so this can be rearranged to the result \begin{eqnarray*} \sum_{j=n +1}^{A_{r}} \frac{1}{j} =r+ 2 \sum_{k=0}^{r-1} (r-k) \sum_{j=A_{k-1}+1}^{A_k} \frac{1}{(3j)^3-3j}. \end{eqnarray*}
• RHS $=$ right hand side. – Donald Splutterwit Oct 29 '17 at 21:55