What is the sum of reciprocals of first $n$ Euclid numbers 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 $e_{n+1}=e_1\dots e_n +1$.
 A: 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 $$e_1\dots e_n = (e_1\dots e_{n-1})e_n = (e_n-1)e_n$$
A: 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
