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?

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    $\begingroup$ This is convergent of the continued fraction starting $[97; 2, 2, 3, 1, 16539, 1, 6, 3]$. The large $16539$ might explain it. $\endgroup$ – joro Feb 26 '16 at 8:51
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    $\begingroup$ Large 16539 is a question, not an answer. $\endgroup$ – Fedor Petrov Feb 26 '16 at 9:20

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: $$ 97.4 + 1/110 = 10715/110 = 2143/22$$

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    $\begingroup$ This is a very nice observation and leads to a question why $10\pi^4-1/11\approx 974.0000012$ is a near integer. $\endgroup$ – Zurab Silagadze Feb 26 '16 at 11:22
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    $\begingroup$ How do you know he looked at patterns in the decimal expansion instead of finding rational approximations in the usual way (as joro said in a comment to the question, $2143/22$ is just a convergent to the continued fraction)? $\endgroup$ – Jeppe Stig Nielsen Feb 26 '16 at 17:23
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    $\begingroup$ I think it's the most natural way. I found very good article : Ramanujan for lowbrows. Berndt and Bhargava discussed about OP and other problems. $\endgroup$ – user68208 Feb 27 '16 at 0:16
  • $\begingroup$ We may want to add another unit fraction to obtain $5\pi^4-\frac{1}{2·11}-\frac{1}{2^9 5^5} \approx 486.99999999955$ $\endgroup$ – Jaume Oliver Lafont Mar 10 '16 at 16:41

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