Euler's Number as Sum of Egyptian Fractions

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Theorem

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

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

This sequence is A006526 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


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:

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


Sources