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This question follows a suggestion by Tito Piezas in Is there an integral or series for $\frac{\pi}{3}-1-\frac{1}{15\sqrt{2}}$?

Q: Is there a series by Ramanujan that justifies the approximation $\pi\approx2\sqrt{1+\sqrt{2}}?$

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Jaume Oliver Lafont
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    Assuming facts not in evidence. Where does it say that there is a series of Ramanujan that justifies that approximation? – Gerry Myerson Jan 30 '16 at 23:36
  • @GerryMyerson: There is in fact an affirmative answer. The question is just to highlight that sometimes there is a reason why simple expressions are relatively close to $\pi$. I'll need time to type the answer. – Tito Piezas III Jan 30 '16 at 23:48
  • I would appreciate if one of moderators would convert this to a comment. Perhaps the original question could be expanded and other relatively simple $\pi$ approximations should be considered? For example, few years ago I found that $$\pi \approx \sqrt{4\cdot e^{1} - 1}.$$ – Alex Feb 01 '16 at 14:34
  • Eliminating $\pi$ between both approximations, one gets $$e\approx\frac{5}{4}+\sqrt{2}$$, with an error about 2%

    Yours gives $$\frac{\sqrt{4e-1}}{\pi}=1.000178...$$

     – Jaume Oliver Lafont Feb 01 '16 at 15:08
  • @Alex: Sorry, but I don't think I'll be able to answer it as a separate question. I looked at my notes and it is harder to find a connection, even if one uses formulas for $\frac{1}{\pi^2}$. – Tito Piezas III Feb 01 '16 at 15:33
  • @Alex $$\pi=\color{green}3.\color{green}1\color{green}4\color{green}1\color{green}5 \color{green}9 \color{green}2\ 65...$$ $$\sqrt{4e-1}=\color{green}3.\color{green}1\color{green}4\ 2 \color{green}1\color{green}5 \color{green} \ 32 \color{green}9\color{green}2\ 5...$$ – Jaume Oliver Lafont Feb 02 '16 at 06:35

1 Answers1

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Part I. One may ask why,

$$\frac{1}{2\pi\sqrt{2}} - \frac{1103}{99^2} \approx 10^{-9}\tag1$$

is such a good approximation? In fact, the convergents of the continued fraction of $\displaystyle\frac{1}{2\pi\sqrt{2}}$ start as,

$$0,\;\frac{1}{8},\;\color{blue}{\frac{1}{3^2}},\;\frac{8}{71},\;\frac{9}{80},\dots\color{blue}{\frac{1103}{99^2}},\dots$$

and it turns out using the 3rd and 9th convergents has the exact formulas,

$$\frac{1}{2\pi\sqrt{2}}-\frac{1}{3^2} = 16\sum_{k=1}^\infty \frac{(4k)!}{k!^4} \frac{1+10k}{(12^4)^{k+1/2}}$$

$$\frac{1}{2\pi\sqrt{2}}-\frac{1103}{99^2} = 16\sum_{k=1}^\infty \frac{(4k)!}{k!^4} \frac{1103+26390k}{(396^4)^{k+1/2}}\tag2$$

so truncating $(2)$ "explains" the approximation $(1)$.

Part II. The OP's post originated with the observation that,

$$\frac{\pi}{3} -\Big(1+\frac{1}{15\sqrt{2}}\Big) \approx 10^{-5}\tag3$$

The convergents of the continued fraction of $\displaystyle\sqrt{2}\left(\frac{\pi}{3}-1\right)$ are,

$$0,\;\frac{1}{14},\;\color{brown}{\frac{1}{15}},\;\frac{55}{824},\dots$$

As there are many Ramanujan-Sato pi formulas that use $\sqrt{2}$, it is possible (though not certain) there is one that is directly responsible for $(3)$.

Part III. Finally, the question. The modest approximation,

$$\pi \approx 2\sqrt{1+\sqrt{2}}\tag4$$

has an explanation as it is the first term of the Ramanujan-type formula,

$$\frac{1}{\pi} - \frac{1}{2\sqrt{1+\sqrt{2}}} = \sum_{n=1}^\infty (-1)^n\frac{(2n)!^3}{n!^6} \frac{16(8+5\sqrt{2})n+8(2+\sqrt{2})}{\big(2^9(1+\sqrt{2})^3\big)^{n+1/2}}\tag5$$

similar to the pair in $(2)$. Truncating the formula immediately implies,

$$\frac{1}{\pi} \approx \frac{1}{2\sqrt{1+\sqrt{2}}}$$

More terms improve the approximation, and an infinity makes it exact.

Tito Piezas III
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  • The coincidence between your unnumbered $\frac{1}{2\pi\sqrt{2}}\approx \frac{1}{9}$ and http://math.stackexchange.com/questions/1623368/is-there-an-integral-or-series-for-frac-pi3-1-frac115-sqrt2#comment3321177_1625142 suggests the additional question whether Ramanujan's series may be obtained from $\left(1+x\right)^\frac{1}{2}asin(\frac{x}{2})$ – Jaume Oliver Lafont Jan 31 '16 at 01:42
  • A "Ramanujan-type" formula isn't the same as a series "by Ramanujan", so this doesn't answer the question that was asked. Don't get me wrong – it's very nice mathematics – it's just not an answer to the question as asked. – Gerry Myerson Feb 01 '16 at 12:07
  • @TitoP Lovely stuff! +1 –  Feb 01 '16 at 14:31
  • @Bacon: Thanks. You'd like formula $(2)$ even more if you find out it is connected to the Pell equation $x^2-29y^2 = 1$. – Tito Piezas III Feb 01 '16 at 14:50
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    @GerryMyerson: In his 1914 paper, Ramanujan gave 17 formulas for four kinds of families. You're right, the particular formula using $1+\sqrt{2}$ isn't explicitly there, but the general series $\displaystyle \frac{1}{\pi}= \sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{An+B}{C^{n+1/2}}$ is by Ramanujan. :) – Tito Piezas III Feb 01 '16 at 14:59
  • Why $\frac{1}{2\pi\sqrt{2}}$ is so close to the fifth convergent $\frac{9}{80} $ is explained by $$\sum_{k=0}^\infty \frac{160}{(8k+5)(8k+6)(8k+7)(8k+9)(8k+10)(8k+11)} =\frac{20}{9}-\frac{\pi}{\sqrt{2}}$$ – Jaume Oliver Lafont Feb 22 '16 at 23:50