The origin of the Ramanujan's $\pi^4\approx 2143/22$ identity What is the origin of the Ramanujan's approximate identity $$\pi^4\approx 2143/22,\;\;\tag 1$$ which is valid with $10^{-9}$ relative accuracy? For comparison, the relative accuracy of the well known $\pi\approx 22/7$ is only $4\cdot10^{-4}$ and in this case we have the identity
$$\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 \frac{dx}{1+x^2}, \tag{2}$$
which explains why the difference is small (concerning this identity, see Source and context of $\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 dx/(1+x^2)$?).
Of course, (1) can be rewritten in the form $$\zeta(4)\approx 2143/1980,$$
so maybe some fast convergent series for $\zeta(4)$ can be used to get this approximate identity (in the case of $\frac{22}{7}-\pi$,  a series counterpart of (2) is $$\sum_{k=0}^\infty \frac{240}{(4k+5)(4k+6)(4k+7)(4k+9)(4k+10)(4k+11)}=\frac{22}{7}-\pi$$ - see Source and context of $\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 dx/(1+x^2)$?).
P.S. I just discovered that this question was discussed in  https://math.stackexchange.com/questions/1359015/is-there-an-integral-for-pi4-frac214322 and in https://math.stackexchange.com/questions/1649890/is-there-a-series-to-show-22-pi42143 is anything to add to the answers given there?
 A: Perhaps the most intuitive way of explaining this identity is with the help of a continued fraction
$$
97+\frac{1}{2+\frac{1}{2+\frac{1}{3+\frac{1}{1+\frac{1}{16539+\frac{1}{\ddots}}}}}}
$$
The unexpectedly large partial quotient of 16539 means that using all partial quotients up to (but not including) this value gives an incredibly close approximation of $\pi^4$ given the number of terms used. What this large number means essentially is that it takes a denominator of that many times more than the previous in order to obtain a better approximation, implying that the previous denominator is a really close approximation. The same method can be used to show that 355/113 is a very close approximation of $\pi$ for the number of partial quotients used.
Evaluating the truncation
$$
97+\frac{1}{2+\frac{1}{2+\frac{1}{3+\frac{1}{1}}}}
$$
gives the famous approximation of $\frac{2143}{22}$ for $\pi^4$.
A: I think Ramanujan's thought was very simple. He calculated the decimal expansion of $\pi^4$ and he got: $$\pi^4 = 97.409091034... \approx 97.4090909...= 97.4 +1/110$$
And then: 
$$ 97.4 + 1/110 = 10715/110 = 2143/22$$
A: FWIW, I found a source according to which

He found this by first squaring the square of π which gives 97.40909103…, then by subtracting 9^2 = 81 he got 16.40909103…, multiplying this by 22 he got 361.0000027…, 361 being the square of 19.

(Thereafter, the source compares to the alternate "method" mentioned in the earlier answer, π^4 = 97.40909103... ≈ 97.409090909...)
The Wikipedia page "Approximations of π" cites his "Lost notebook page 16"

from Ramanujan, who claimed the Goddess of Namagiri appeared to him in a dream and told him the true value of π.

