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
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $0 \cdotp 36787 \, 94411 \, 71442 \, 32159 \, 55237 \, 70161 \, 46086 \, 74458 \, 11131 \, 031$