# Chaitanya's Random Pages

## November 22, 2014

### A cute sum of Ramanujan

Filed under: mathematics — ckrao @ 3:09 am

Here is a beautiful sum I found in , apparently due to Ramanujan. $\displaystyle \frac{1}{1^3\times 2^3} + \frac{1}{2^3 \times 3^3} + \frac{1}{3^3 \times 4^3} + \cdots = \sum_{k=1}^{\infty} \frac{1}{k^3(k+1)^3} = 10-\pi^2\quad\quad(1)$

Note that the result also demonstrates that $\pi^2$ is slightly less than 10, an alternative to the approaches in .

Many of his results require advanced number theory to prove, but this one is not too tricky, as long as we know the following similarly attractive result that I had previously mentioned in this blog post. $\displaystyle \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}\quad\quad(2)$

To prove (1), the idea is to write $\frac{1}{k^3(k+1)^3}$ as a sum of partial fractions and then sum a telescoping series. You might wish to try this yourself before reading further.

We write $\displaystyle \frac{1}{k^3(k+1)^3} = \frac{a(k)}{k^3} + \frac{b(k)}{(k+1)^3} = \frac{a(k)(k+1)^3 + b(k)k^3}{k^3(k+1)^3},\quad\quad(3)$

where $a(k)$ and $b(k)$ are quadratic polynomials. One way of finding $a(k)$ and $b(k)$ would be to compare coefficients of 1, $k$, $k^2$, $k^3$ and solve a system of equations. Another approach, the Extended Euclidean Algorithm, does so via finding the greatest common divisor of $k^3$ and $(k+1)^3$.

The following manipulations verify that the greatest common divisor of $k^3$ and $(k+1)^3$ is 1, where the next line computes quotients and remainders based on the previous line. \displaystyle \begin{aligned} (k+1)^3 &= k^3 + (3k^2 + 3k + 1) & \quad (4)\\ k^3 &= \frac{k}{3}(3k^2 + 3k + 1) - \left(k^2 + \frac{k}{3}\right) & \quad (5)\\ 3k^2 + 3k+ 1 &= 3\left(k^2 + \frac{k}{3}\right) + (2k + 1) & \quad (6)\\ k^2 + \frac{k}{3} &= \frac{k}{2}(2k+1) - \frac{k}{6} & \quad (7)\\ 2k+1 &= 12 \frac{k}{6} + 1 & \quad (8) \end{aligned}

Reversing the steps of (4)-(8) gives us 1 as quadratic polynomial combinations of $k^3$ and $(k+1)^3$, thus providing us with $a(k)$ and $b(k)$. \displaystyle \begin{aligned} 1 &= (2k+1) - 12\left[\frac{k}{2}(2k+1) - \left(k^2 + \frac{k}{3}\right)\right]\\ &= (2k+1)(1-6k) + 12\left(k^2 + \frac{k}{3}\right)\\ &= \left[3k^2 + 3k + 1 - 3\left(k^2 + \frac{k}{3}\right) \right](1-6k) + 12\left(k^2 + \frac{k}{3}\right)\\ &= (3k^2 + 3k+1)(1-6k) + \left[12 - 3(1-6k)\right]\left(k^2 + \frac{k}{3}\right)\\ &= (3k^2 + 3k+1)(1-6k) + \left[12 - 3(1-6k)\right]\left[\frac{k}{3}(3k^2+3k+1) - k^3\right]\\ &= (3k^2 + 3k + 1)\left[1-6k + 4k - k(1-6k)]\right] + \left[12 - 3(1-6k)\right]\left(-k^3\right)\\ &= (k+1)^3\left(6k^2 - 3k+ 1\right) - k^3\left(6k^2 - 3k+ 1 + 12 - 3 + 18k\right)\\ &= (k+1)^3(6k^2 - 3k+ 1) - k^3(6k^2 + 15k+10).\quad\quad(9) \end{aligned}

Using (9) we then have \displaystyle \begin{aligned} \sum_{k=1}^{\infty} \frac{1}{k^3(k+1)^3} &= \lim_{N \rightarrow \infty} \sum_{k=1}^{N} \frac{1}{k^3(k+1)^3}\\ &= \lim_{N \rightarrow \infty} \sum_{k=1}^{N} \frac{ (k+1)^3(6k^2 - 3k+ 1) - k^3(6k^2 + 15k+10)}{k^3(k+1)^3}\\ &= \lim_{N \rightarrow \infty} \left(\sum_{k=1}^{N} \frac{ 6k^2 - 3k+ 1}{k^3} - \frac{6k^2 + 15k+10}{(k+1)^3}\right)\\ &= \lim_{N \rightarrow \infty} \sum_{k=1}^{N} \left(\frac{ 6k^2 - 3k+ 1}{k^3} - \frac{6(k+1)^2 + 3(k+1) + 1}{(k+1)^3}\right)\\ &= \lim_{N \rightarrow \infty} \sum_{k=1}^{N} \left(\frac{6}{k} - \frac{3}{k^2} + \frac{1}{k^3}\right) - \sum_{k=2}^{N+1} \left(\frac{6}{k} + \frac{3}{k^2} + \frac{1}{k^3}\right)\\ &= \left(\frac{6}{1} - \frac{3}{1} + \frac{1}{1}\right) - \lim_{N \rightarrow \infty} \sum_{k=2}^{N} \frac{6}{k^2} - \lim_{N \rightarrow \infty}\left(\frac{6}{N+1} + \frac{3}{(N+1)^2} + \frac{1}{(N+1)^3} \right)\\ &= 4 - (\pi^2 - 6) - 0 \quad\text{(using (2))}\\ &= 10-\pi^2, \end{aligned}

thus verifying (1).

The interested reader might like to find other identities in a similar manner. 🙂

#### References

 Clifford A. Pickover, A Passion for Mathematics: Numbers, Puzzles, Madness, Religion, and the Quest for Reality, John Wiley & Sons, 2005.