Square Root of 2 as Sum of Egyptian Fractions

Theorem
The square root of $2$ can be approximated by the following sequence of Egyptian fractions:


 * $\sqrt 2 = 1 + \dfrac 1 3 + \dfrac 1 {13} + \dfrac 1 {253} + \dfrac 1 {218 \, 201} + \dfrac 1 {61 \, 323 \, 543 \, 802} + \cdots$

Proof
We have by definition of the square root of $2$ that:
 * $\sqrt 2 - 1 \approx 0 \cdotp 41421 \, 35623 \, 73095 \, 04880 \ldots$

By inspection:
 * $\dfrac 1 3 < \sqrt 2 - 1 < \dfrac 1 2$

Thus:
 * $\sqrt 2 - 1 - \dfrac 1 3 \approx 0 \cdotp 08088 \, 02290 \, 39761 \, 71546 \ldots$

Then:
 * $\dfrac 1 {13} = 0 \cdotp 07692 \, 3 \ldots$ repeating

and:
 * $\dfrac 1 {12} = 0 \cdotp 08333 \ldots$ repeating

and so:
 * $\sqrt 2 - 1 - \dfrac 1 3 - \dfrac 1 {13} \approx 0 \cdotp 00395 \, 71521 \, 16684 \, 79239 \cdots$

Thus one can generate this sequence of denominators $\sequence {D_n}$ by:


 * $\ds D_n = \ceiling {\paren {\sqrt 2 - \sum_{i \mathop = 0}^{n - 1} \frac 1 {D_i}}^{-1}}$