Euler's Number as Sum of Egyptian Fractions

Theorem
The reciprocal of Euler's number $e$ can be approximated by the following sequence of Egyptian fractions:


 * $\dfrac 1 e = \dfrac 1 3 + \dfrac 1 {29} + \dfrac 1 {15 \, 786} + \dfrac 1 {513 \, 429 \, 610} + \cdots$

Proof
We have by definition of the reciprocal of $e$ that:
 * $\dfrac 1 e \approx 0 \cdotp 36787 \, 94411 \, 71442 \, 32159 \, 55237 \, 70161 \, 46086 \, 74458 \, 11131 \, 031 \ldots$

By inspection:
 * $\dfrac 1 3 < \dfrac 1 e < \dfrac 1 2$

Thus:
 * $\dfrac 1 e - \dfrac 1 3 \approx 0 \cdotp 03454 \, 61078 \, 38109 \, 08826 \, 21904 \, 36828 \, 12753 \, 41124 \, 77797 \, 70 \ldots$

Then:
 * $\dfrac 1 {29} = 0 \cdotp 03448 \, 27586 \, 20689 \, 65517 \, 24137 \, 931 \ldots$ repeating

and:
 * $\dfrac 1 {28} = 0 \cdotp 03571 \, 42857 \, 14285 \ldots$ repeating

and so:
 * $\dfrac 1 e - \dfrac 1 3 - \dfrac 1 {29} \approx 0 \cdotp 00006 \, 33482 \, 17499 \, 43308 \, 97766 \, 33724 \, 67925 \, 82504 \, 08832 \, 19 \cdots$

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


 * $D_n = \ceiling {\paren {\dfrac 1 e - \displaystyle \sum_{i \mathop = 0}^{n - 1} \frac 1 {D_i}}^{-1}}$