A series related to $\pi\approx 2\sqrt{1+\sqrt{2}}$

Question

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 $\pi\approx2\sqrt{1+\sqrt{2}}?$

Answer 1

The answer needs some context.

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,

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

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

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

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

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 $\displaystyle\sqrt{2}\big(\frac{\pi}{3}-1\big)$ are,

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

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

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 (after some manipulation) of the Ramanujan-type formula,

$$\frac{1}{\pi}= \sum_{n=0}^\infty (-1)^n\frac{(2n)!^3}{n!^6} \frac{16(8+5\sqrt{2})n+8(2+\sqrt{2})}{(C)^{n+1/2}},\quad C = 2^9(1+\sqrt{2})^3\tag5$$

And there are many other similar ones.