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$S = \frac{1}{q+1}+\frac{1}{(q+2)(q+1)}+\frac{1}{(q+3)(q+2)(q+1)}...$

I know that $0<S<1$. But is it rational?

I took this series from(proof by contradiction that $e$ is is irrational ): http://www.mathshelper.co.uk/Proof%20That%20e%20Is%20Irrational.pdf

In the paper it assumes that $e=p/q$. So $q$ is not allowed to be $0$.

Pinteco
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4 Answers4

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It is irrational in general for positive integer $q$.

This sum represents the Engel expansion of some number. Engel expansion is unique, and it is finite if and only if the number is rational.

(If every $a_k$ is different though!)

Since in this case the expansion is infinite by definition, the number is irrational.

Yuriy S
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  • new information +1 – Shobhit Feb 15 '18 at 20:05
  • @Shobhit, yes, it's pretty neat. Though I added the condition of every factor $a_k$ being distinct, but in this case it holds. Usually Engel expansion is defined with each factor unique, but apparently someone decided to add the part about 'infinite Engel expansion for the rationals' to the Wikipedia article (i.e. geometric series) and scared me – Yuriy S Feb 15 '18 at 20:12
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\begin{eqnarray*} \underbrace{e}_{\text{irrational}}=\underbrace{1+\frac{1}{1!}+ \cdots+ \frac{1}{q!}}_{rational} +\frac{1}{q!}\left( \frac{1}{q+1} +\frac{1}{(q+1)(q+2)}+\cdots \right) \end{eqnarray*} so ...

Donald Splutterwit
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I doubt $S(q)$ be rational; in particular $S(0)=e-1$ which is trascendental. Actually $$S(q)=\begin{eqnarray*} e=1+\frac{1}{1!}+ \cdots+ \frac{1}{q!} +\frac{1}{q!}\left( \frac{1}{q+1} +\frac{1}{(q+1)(q+2)}+\cdots \right) \end{eqnarray*}$$ Consequently $S(q)$ is always trascendental because of it is equal to $e$ minus a rational.

Piquito
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If this sum be rational, then according to the paper you linked above, the number $e$ will be a rational number.

Qurultay
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