How to prove this question by Ramanujan? click here for photo
$$1+2\sum_{k=1}^\infty  \frac{1}{(4k)^3-(4k)}= \frac{3}{2}\ln(2)\,.$$
well i have attatched a photo which has been asked to prove without using calculus,but how to solve this using calculus ?
 A: I have no clue how to solve this without calculus.  At least, I do not know how to define the natural logarithm without calculus.  I am borrowing some part of this solution from mechanodroid's deleted solution.  
First, write
$$\frac{1}{(4k)^3-(4k)}=-\frac{1}{4k}+\frac{1}{2(4k-1)}+\frac{1}{2(4k+1)}$$
for every positive integer $k$.  Define
$$S_n:=1+2\,\sum_{k=1}^n\,\frac{1}{(4k)^3-(4k)}$$
for all $n=1,2,3,\ldots$.  Hence, the $n$-th partial sum is given by $$
\begin{align}
S_{n}&=1+\sum_{k=1}^{n}\,\left(-\frac{1}{2k}+\frac{1}{4k-1}+\frac{1}{4k+1}\right)
\\&=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots-\frac{1}{2n}\right)+\left(\frac{1}{2n+1}+\frac{1}{2n+3}+\ldots+\frac{1}{4n+1}\right)
\\&=T_n+U_n\,,
\end{align}$$
where $T_n:=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots-\frac{1}{2n}$ and $U_n:=\frac{1}{2n+1}+\frac{1}{2n+3}+\ldots+\frac{1}{4n+1}$.  It remains to show that $$\lim_{n\to\infty}\,T_n=\ln(2)\text{ and }\lim_{n\to\infty}\,U_n=\frac{1}{2}\,\ln(2)\,.$$
The former is well known, whilst the latter follows from the fact that $2U_n$ is a Riemann sum for $\int_{2}^{4}\,\frac{1}{x}\,\text{d}x$.  Alternatively, observe that $T_n=\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}$, and
$$\frac{1}{2} \,T_{n+1}\leq U_n\leq \frac{1}{2} T_{n+1}+\frac{1}{2n+1}-\frac{1}{4n+2}\,.$$
