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This is a practice question for my exam tomorrow: Let

$$I_n = \int_0^1 e^tt^ndt$$ where $n$ is a non-negative integer. Assume the following results:

$$I_{n+1} = e - (n + 1)I_n$$ $$I_n = (-1)^{n + 1}n! + e\sum_{r = 0}^n (-1)^r\frac{n!}{(n-r)!}$$ $$\frac{1}{n+1}\le I_n < \frac{e}{n} ~~~\text{for all $n \ge 1$}$$

Use the above results to prove that $e$ is irrational.

The only thing I can think of is to use the pinching theorem on the last result to argue that

$$I_\infty = 0$$

as $n \to \infty$.

I'm more or less caught in a pinch afterwards (heh).

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  • $\begingroup$ If $e$ were rational, what would the denominator of $I_n$ be, according to result #2? $\endgroup$ – ziggurism Jul 10 '14 at 13:44
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Assume $e=a/b$ for $a,b \in \mathbb{N}$.

Then, multiplying your second equation by $b$, that would imply that $I_n b\in \mathbb{N}$ for all $n$ (the RHS is an integer).

But taking your third equation, and picking $n=a$, we'd get $I_n b < 1$. Contradiction.

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