We know about Euclid Number . I want to know the sum of reciprocals of 1st $n$ Euclid Number ? In that book I have been told to find the value of following series :

$$ \dfrac{1}{e_1}+\dfrac{1}{e_2}+\cdots+\dfrac{1}{e_n} = ? $$

In fact, I can not understand the following calculations: $$ \dfrac{1}{e_1}+\dfrac{1}{e_2}+\cdots+\dfrac{1}{e_n} = 1 - \dfrac{1}{e_n(e_{n}-1)} = 1-\dfrac{1}{e_{n+1}-1}$$

Can you please help me to find how this sum is done? This math is taken from "Concrete Math" Of Ronald L. Graham, Donald E. Knuth, and Oren Patashnik.

The definition in this book for Euclid numbers is non-standard: $e_1=2$ and $e_{n+1}=e_1\dots e_n +1$.

  • 2
    $\begingroup$ The usual Euclid numbers are $p_0p_1\cdots p_n+1$ where the $p_i$ are the first $n+1$ primes. This must be some other Euclid. $\endgroup$ – André Nicolas Dec 17 '12 at 21:36
  • $\begingroup$ Where about in the book is this statement? $\endgroup$ – Thomas Russell Dec 17 '12 at 21:53
  • 1
    $\begingroup$ On page 108 of Concrete Mathematics, it suggests we define the Euclid numbers by the recurrence: $$e_{n}=e_{1}e_{2}\dots e_{n-1} + 1$$ It then goes on to state that there's a constant $E\approx1.264$, such that: $$e_{n}=\left\lfloor E^{2^{n}}+\frac{1}{2}\right\rfloor$$ $\endgroup$ – Thomas Russell Dec 17 '12 at 22:03
  • $\begingroup$ And $e_1=2$ to complete the definition. $\endgroup$ – Ross Millikan Dec 17 '12 at 22:08
  • $\begingroup$ @RossMillikan Yes, sorry, I forgot to include that!! $\endgroup$ – Thomas Russell Dec 17 '12 at 22:11

Given $e_1=2$ and $e_{n+1}=e_1\dots e_n+1$, then:

$$\frac{1}{e_1}+...+\frac 1 {e_n} = 1-\frac 1 {e_1\dots e_n} = 1-\frac 1 {e_{n+1}-1}$$

This can easily be proved by induction. Just compute:

$$\frac{1}{e_1}+...+\frac 1 {e_n} + \frac{1}{e_{n+1}} = 1-\frac 1 {e_{n+1}-1} + \frac{1}{e_{n+1}}$$

Just multiply it out and used that $e_{n+2}-1 = e_{n+1}(e_{n+1}-1)$.

You get the middle term of your question by noting that $$e_1\dots e_n = (e_1\dots e_{n-1})e_n = (e_n-1)e_n$$

  • 2
    $\begingroup$ The book actually states that: $$\frac{1}{e_{1}}+\frac{1}{e_{2}}+\cdots+\frac{1}{e_{n}}=1-\frac{1}{e_{n}(e_{n}-1)}=1-\frac{1}{e_{n+1}-1}$$ $\endgroup$ – Thomas Russell Dec 17 '12 at 22:14
  • $\begingroup$ Plz see the edited version . I do not want the proof . $\endgroup$ – Way to infinity Dec 19 '12 at 15:45
  • $\begingroup$ What do you want, then? A hint for the proof? The hint is: Induction. @SultanAhmedSagor $\endgroup$ – Thomas Andrews Dec 19 '12 at 15:59

I don't know what your right hand sides are supposed to be, but $\frac 13 + \frac 17=\frac {10}{21}$ and $\frac 13 + \frac 17+\frac 1{31}=\frac{331}{651}$ which do seem to be of the form you list. They seem to be much farther from $1$.

Added: the Euler numbers you are using are defined by $e_1=2, e_n=\prod_{i=1}^{n-1} e_i +1$. They start $2,3,7,43,1807,\ldots $ and are given in A000058 of OEIS as Sylvester's sequence. The proof you are looking for is given under Connection with Egyptian fractions in the Wikipedia article


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