How can I prove that $$e^n \neq \sum_{i=0, i\neq n }^{+\infty}k_ie^{i} $$ where e is the base of the natural logarithm, $k_i$ are non-negative integers, and $n$ is a positive integer?
Asked
Active
Viewed 47 times
1
-
@QiaochuYuan That's what I thought initially, but I'm not sure how to rule out the case that the sum goes infinitely to the left. – Aaron Jan 20 '18 at 07:15
-
I made a mistake in the statement, there is no infinite sum to the left, it starts at zero. Still it is not clear to me how to prove it. – Sunny88 Jan 20 '18 at 07:18
-
@Sunny88 Suppose you had such an expression. Then $e$ is the root of a polynomial with rational coefficients. But $e$ cannot be the root of a polynomial with rational coefficients (because it is transcendental). – Aaron Jan 20 '18 at 07:22
-
@Sunny88: I deleted my comment after reading that, but actually that was unnecessary: in order for this sum to converge it must be finite, since otherwise it fails the limit test. So we are back to transcendentality. – Qiaochu Yuan Jan 20 '18 at 07:24