This is a problem that often appears
in various contexts:
Compare the magnitudes of n^{n+1} and (n+1)^n (where n is non-negative).
In elementary school, we use direct
calculation;
In middle school, we look for patterns from
examples;
In high school, we proceed directly to the
proof.
The answer to this problem is: when n > e, we have n^{n+1} > (n+1)^n.
I provide two proofs:
Proof 1:
To prove n^{n+1} > (n+1)^n is equivalent to proving: n > \frac{(n+1)^n}{n^n} = \left(1 + \frac{1}{n}\right)^n
From the study of the constant e, we know that for non-negative n, the expression (1 + 1/n)^n is monotonically increasing. Since \lim_{n \to +\infty} (1 + 1/n)^n = e, it follows that (1 + 1/n)^n < e. Therefore, when n > e, the inequality n > \frac{(n+1)^n}{n^n} = (1 + 1/n)^n holds. Q.E.D.
Proof 2:
This proof is relatively more general and can be used to compare n^{n+m} and (n+m)^n, among other cases. Here we only discuss the case where m=1.
Consider the derivative of the function f(x) = \frac{\ln x}{x}: \left(\frac{\ln x}{x}\right)' = \frac{1 - \ln x}{x^2} When x > e, we have \frac{1 - \ln x}{x^2} < 0, which means the function \frac{\ln x}{x} is monotonically decreasing in this interval. If e < a < b, then: \frac{\ln a}{a} > \frac{\ln b}{b} \iff \frac{\ln a}{\ln b} > \frac{a}{b}
And when n > e: n^{n+1} = (n+1)^{(n+1) \cdot \frac{\ln n}{\ln(n+1)}} > (n+1)^{(n+1) \cdot \frac{n}{n+1}} = (n+1)^n
Q.E.D.
When reposting, please include the original link: https://kexue.fm/archives/132
For more details regarding reposting, please refer to: “Scientific Space FAQ”