Here is a beautiful sum I found in [1], apparently due to Ramanujan.
Note that the result also demonstrates that is slightly less than 10, an alternative to the approaches in [2].
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.
To prove (1), the idea is to write as a sum of partial fractions and then sum a telescoping series. You might wish to try this yourself before reading further.
We write
where and are quadratic polynomials. One way of finding and would be to compare coefficients of 1, , , and solve a system of equations. Another approach, the Extended Euclidean Algorithm, does so via finding the greatest common divisor of and .
The following manipulations verify that the greatest common divisor of and is 1, where the next line computes quotients and remainders based on the previous line.
Reversing the steps of (4)-(8) gives us 1 as quadratic polynomial combinations of and , thus providing us with and .
Using (9) we then have
thus verifying (1).
The interested reader might like to find other identities in a similar manner. 🙂
References
[1] Clifford A. Pickover, A Passion for Mathematics: Numbers, Puzzles, Madness, Religion, and the Quest for Reality, John Wiley & Sons, 2005.
[2] Noam D. Elkies, Why is so close to 10?
A better way to decompose this expression is with derivatives
1/(k*(k+1))^3 = A/k + B/k^2 + C/k^3 + D/(k+1) + E/(k+1)^2 + F/(k+1)^3
(k+1)^-3 = (A*k^2 + B*k + C) + k^3 * whatever
-3*(k+1)^-4 = (2A*k + B) + k^2 * whatever // d/dk
12*(k+1)^-5 = (2A) + k * whatever // d^2/dk^2
k=0 -> A = 6, B = -3, C = 1
Do the same for D,E,F, we have D = -6, E = -3, F = -1
1/(k*(k+1))^3 = 6*(1/k-1/(k+1)) – 3*(1/k^2+1/(k+1)^2) + (1/k^3-1/(k+1)^3)
Σ(1/(k*(k+1))^3, k=1..inf) = 6*1 – 3*(pi^2/6 + pi^2/6-1) + 1 = 10 – pi^2
Comment by Albert Chan — October 26, 2021 @ 12:14 am |