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Let $\{\epsilon_n\}$ be a sequence where $\epsilon_n$ is either $ 1$ or $-1$. How could I Show
that the sum of the series

$$\sum_{n=0}^{\infty}\frac{\epsilon_n}{n!}$$

is an irrational number.

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Just repeat the arguments in the regular proof that e is irrational. –  achille hui May 9 '13 at 15:17
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2 Answers 2

If there exist $p\in\mathbb{Z}$ and $q\in\mathbb{N}$, such that $\frac{p}{q}=\sum_{n=0}^\infty\frac{\epsilon_n}{n!}$, then $q!\cdot\sum_{n=q+1}^\infty\frac{\epsilon_n}{n!}$ must be an integer. However, $$|q!\cdot\sum_{n=q+1}^\infty\frac{\epsilon_n}{n!}-\frac{\epsilon_{q+1}}{q+1}|\le\sum_{n=q+2}\frac{q!}{n!}<\frac{1}{(q+1)(q+2)}\cdot\sum_{m=0}\frac{1}{2^m}=\frac{2}{(q+1)(q+2)},$$ which implies that $$0<\frac{1}{q+1}-\frac{2}{(q+1)(q+2)}\le |q!\cdot\sum_{n=q+1}^\infty\frac{\epsilon_n}{n!}|\le \frac{1}{q+1}+\frac{2}{(q+1)(q+2)}<1,$$ a contradiction.

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The proof is by contradiction. Define: $$ S = \sum_{k \ge 0} \frac{\epsilon_k}{n!} $$ Assume $S$ is rational, i.e. there are $u \in \mathbb{Z}$, $v \in \mathbb{N}$ such that $S = u / v$.

Pick $b > v$, so that $b \ge 2$. Then $b! S$ is an integer, i.e.: $$ S = \sum_{0 \le k \le b} \frac{b! \epsilon_k}{k!} + \sum_{k > b} \frac{b! \epsilon_k}{k!} $$ The first sum is an integer, so the second sum has to be an integer too. Now: $$ \frac{b! \epsilon_k}{k!} = \frac{\epsilon_k}{(b + 1) (b + 2) \ldots k} $$ But: $$ \frac{1}{(b + 1) (b + 2) \ldots k} < \frac{1}{b^{k - b}} $$ By the triangular inequality: $$ \left\rvert \sum_{k \ge b + 1} \frac{b! \epsilon_k}{k!} \right\rvert \le \sum_{k \ge b + 1} \frac{b!}{k!} < \sum_{k \ge b + 1} b^{-k} = b^{- b - 1} \sum_{k \ge 0} b^{-k} = b^{- b - 1} \frac{1}{1 - 1 / b} = \frac{1}{b^{b +1} (b - 1)} < \frac{1}{b^{b + 1}} $$ So: $$ 1 = \left\lvert \frac{b! \epsilon_b}{b!} \right\rvert > \left\rvert \sum_{k \ge b + 1} \frac{b! \epsilon_k}{k!} \right\rvert $$ and the "leftover sum" can never be 0, so it isn't an integer.

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