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Let $S=\{m\cdot n^r\mid m,n\in\mathbb Z,r\in\mathbb Q\}$
Can $e$ or $\pi$ be written as a finite sum of elements of $S$?

Can $\pi=xe$, with $x$ algebraic?

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Since the members of $S$ are algebraic, while $e$ and $\pi$ are transcendental... – J. M. Nov 27 '11 at 13:17
Also noting that a finite sum of algebraics is algebraic... – J. M. Nov 27 '11 at 13:21
For the new question: it is not yet known if $\pi/e$ is algebraic or transcendental. – J. M. Nov 27 '11 at 13:26
See also this thread. – J. M. Nov 27 '11 at 13:31
up vote 3 down vote accepted

As finite sums, no. If you could, you'll get that $\pi$ and $e$ are algebraic numbers, which is known to be false.

As infinite sums:

$$e = \sum_{n=0}^{\infty} \frac1{n!}$$

$$\pi = 4\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}$$

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