[ocfnash]

Weirdly, I was just sent this same blog post a few days ago. I guess it’s doing the rounds. I’ll repeat what I said to my friend who sent it.

Ramanujan was of course an extremely talented mathematician but IMHO, there is an unnecessary cult of mystery about him. In particular, there is no mystery as to where this formula comes from: it is obtained from applying a standard technique to a standard formula. Also, it is just one instance, I’ll call it the “k = 6” instance, of a family of such formulae and I’m near-certain the “k = 2” case was known before Ramanujan (on phone or would double check).

In any case, the standard formula is just the series expansion for: n! / sqrt{2πn} (n/e)^n and the standard technique is to raise both sides of a series expansion to some power, k say, multiply out one side, and then take k-th roots again.

In this case we just need to calculate: (1 + 1/12n + 1/288n^2 - 139/51840n^3 + O(1/n^4))^6 = 1 + 1/2n + 1/8n^2 + 1/240n^3 + O(1/n^4)

After taking the 6th root again we multiply inside by 8n^3 (from the LHS) and you get Ramanujan’s formula.

[sanjayio]

This might be obvious with a modern mathematical education but keep in mind he was born in 1887 and died at the age of 32. Theorems are a little like magic tricks, they become very apparent once someone tells you the trick.

[pfortuny]

In this case it looks a bit obvious: Stirling is to degree 1 in the square root, “Ramanujan”’s to degree 4. Nothing to see here.