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).

  • $\begingroup$ If $e$ were rational, what would the denominator of $I_n$ be, according to result #2? $\endgroup$ – ziggurism Jul 10 '14 at 13:44

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.