How to prove this question by Ramanujan?

Question

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 ?

Answer 1

Note that we have from partial fraction expansion

$$\begin{align} \sum_{k=1}^{K}\frac{1}{(4k)^3-(4k)}&=\sum_{k=1}^{K}\left(-\frac1{4k}+\frac{1/2}{4k-1}+\frac{1/2}{4k+1}\right)\tag1 \end{align}$$

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Next, we make use of "ungrouping" even and odd terms of the sum $\sum_{k=1}^K\left(\frac{1}{4k+1}+\frac{1}{4k-1}\right)$ to write

$$\begin{align} \sum_{k=1}^K\left(\color{blue}{\frac{1}{4k+1}}+\color{red}{\frac{1}{4k-1}}\right)&=\sum_{k=1}^K\left(\color{blue}{\frac{1}{2\underbrace{(2k)}_{\text{even terms}}+1}}+\color{red}{\frac{1}{2\underbrace{(2k-1)}_{\text{odd terms}}+1}}\right)\\\\ &=\sum_{k=1}^{2K}\frac1{2k+1}\tag2 \end{align}$$

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Substituting $(2)$ into $(1)$, we obtain

$$\begin{align} \sum_{k=1}^{K}\frac{1}{(4k)^3-(4k)}&=\frac12\left(\sum_{k=1}^{2K}\color{purple}{\frac{1}{2k+1}}-\sum_{k=1}^{K}\color{orange}{\frac1{2k}}\right)\\\\ &=\frac12\sum_{k=1}^{2K}\left(\color{purple}{\frac{1}{2k+1}}-\color{orange}{\frac1{2k}}\right)+\frac12\sum_{k=K+1}^{2K}\color{orange}{\frac1{2k}}\tag3 \end{align}$$

We can proceed two distinct ways to continue from $(3)$. The first relies on the power series definition of the logarithm function while the second relies on the integral definition of the logarithm function and Riemann Sums.

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METODOLOGY $1$: Using only $\displaystyle \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}=\log(2)\tag4$

Here, we note that the general terms of the first summation on the right-hand side of $(3)$ is a grouping of the difference of successive even and odd reciprocal integers. Hence, we can be expressed that summation as

$$\sum_{k=1}^{2K}\left(\frac{1}{2k+1}-\frac1{2k}\right)=\sum_{k=2}^{4K+1}\frac{(-1)^{k-1}}{k}\tag5$$

Using the result in $(5)$, the second summation on the right-hand side of $(3)$ can be written as

$$\begin{align} \sum_{k=K+1}^{2K}\frac1{2k}&=\frac12\sum_{k=K+1}^{2K}\frac1k\\\\ &=\frac12 \color{magenta}{\sum_{k=1}^{2K}\frac1k} -\frac12\sum_{k=1}^K\frac1k\\\\ &=\frac12 \color{magenta}{\sum_{k=1}^K \left(\frac1{2k-1}+\frac{1}{2k}\right)}-\frac12\sum_{k=1}^K\frac1k\\\\ &=\frac12 \sum_{k=1}^K \left(\frac1{2k-1}-\frac{1}{2k}\right)\\\\ &=\frac12\sum_{k=1}^K \frac{(-1)^{k-1}}{k}\tag6 \end{align}$$

Substitution of $(5)$ and $(6)$ into $(3)$, letting $K\to \infty$, and exploiting $(4)$ yields the coveted result

$$\begin{align} \bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^{\infty}\frac{1}{(4k)^3-(4k)}=\frac34\log(2)-\frac12} \end{align}$$

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METODOLOGY $2$: Using only Riemann Sums

Here, we write the firsts summation on the right-hand side of $(3)$ as

$$\begin{align} \sum_{k=1}^{2K}\left(\frac{1}{2k+1}-\frac1{2k}\right)&=\color{aqua}{\sum_{k=1}^{2K}\left(\frac{1}{2k+1}+\frac1{2k}\right)}-\color{fuchsia}{2\sum_{k=1}^{2K}\frac1{2k}}\\\\ &=\color{aqua}{\sum_{k=2}^{4K+1}\frac1k}-\color{fuchsia}{\sum_{k=1}^{2K}\frac1{k}}\\\\ &=-1+\sum_{k=2K+1}^{4K+1}\frac1k\\\\ &=-1+\sum_{k=1}^{2K+1}\frac{1}{k+2K}\\\\ &=-1+\frac1K\sum_{k=1}^{2K+1}\frac{1}{2+(k/K)}\tag7 \end{align}$$

Similarly, we can write the second summation on the right-hand side of $(3)$ as

$$\sum_{k=K+1}^{2K}=\frac12 \frac1K\sum_{k=1}^K\frac{1}{1+(k/K)}\tag8$$

Substituting $(7)$ and $(8)$ into $(3)$ reveals

$$\sum_{k=1}^{K}\frac{1}{(4k)^3-(4k)}=\frac12\left(-1+\frac1K\sum_{k=1}^{2K+1}\frac{1}{2+(k/K)} +\frac12 \frac1K\sum_{k=1}^K\frac{1}{1+(k/K)}\right)\tag9$$

Recognizing that the first summation on the right-hand side of $(9)$ is a Riemann Sum for $\int_0^2 \frac{1}{2+x}\,dx=\log(2)$ and that the second summation on the right-hand side of $(9)$ is a Riemann Sum for $\int_0^1 \frac{1}{1+x}\,dx=\log(2)$, we obtain upon letting $K\to \infty$

$$\begin{align} \bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^{\infty}\frac{1}{(4k)^3-(4k)}=\frac34\log(2)-\frac12} \end{align}$$

as expected!

Answer 2

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 $$\frac{1}{2} \,T_{n+1}\leq U_n\leq \frac{1}{2} T_{n+1}+\frac{1}{2n+1}-\frac{1}{4n+2}\,.$$