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