Deriving $\sqrt2 \approx 1 + \frac13 + \frac1{3 \cdot 4} - \frac1{3 \cdot 4 \cdot34}$

Question

Here is a wierd expansion for $\sqrt2$ found in the ancient Indian mathematical literature. $$1 + \frac13 + \frac1{3 \cdot 4} - \frac1{3 \cdot 4 \cdot34} = \frac {577}{408}$$ Today we know that the resulting fraction can be obtained using Pell numbers i.e. the recursion $\frac{P_{n-1}}{P_n} - 1$ .

Can someone explain how can we come up with that particular expansion?

Answer 1

EDIT: This is an answer for an algorithm to generate the unit fractions of the expansion : $$\sqrt2 \approx 1 + \frac12 - \frac1{3 \cdot 4} - \frac1{3 \cdot 4 \cdot34}$$

(an answer to the actual question is provided by Gerry Myerson in his first comment) $$-$$

This (signed) Egyptian fraction may be obtained by starting with the 'exact' $\sqrt{2}$ and removing at each iteration the multiplicative inverse of the nearest integer of the remainder : \begin{array} {c|cc} x&1/x&[1/x]\\ \sqrt{2}&0.707106781187&1\\ \sqrt{2}-1&2.41421356237&2\\ \sqrt{2}-1-\frac 12&-11.6568542495&-12\\ \sqrt{2}-1-\frac 12+\frac 1{12}&-407.646752982&-408\\ \end{array}

This method may be generalized at wish...