A problem that could show up anywhere from elementary school to high school

一道从小学到高中都可能考到的题目 苏剑林 · 2009-09-20 · Mathematics proof inequality exponent logarithm

This is a problem that comes up quite often, in various forms:

Compare the sizes of $n^{n+1}$ and $(n+1)^n$ (where n is non-negative).

In elementary school we'd tackle this by direct computation;

in middle school we'd look at some examples and try to spot a pattern;

in high school we'd go ahead and prove it directly. more

The answer to this problem is: when n>e, we have $n^{n+1}>(n+1)^n$.

I'll give two proofs.

Proof one:

To prove $n^{n+1}>(n+1)^n$, it suffices to prove $n>\frac{(n+1)^n}{n^n}=(1+1/n)^n$.

Now, from what we know about the study of e, for non-negative n, $(1+1/n)^n$ is monotonically increasing; and $\lim_{x->+\infty}(1+1/n)^n=e$, so $(1+1/n)^n < e$. Hence when n>e, $n>\frac{(n+1)^n}{n^n}=(1+1/n)^n$ holds. Q.E.D.

Proof two:

This proof is somewhat more general, and can be used to compare cases like $n^{n+m}$ and $(n+m)^n$. Here we only discuss the case m=1.

Consider the function $(\frac{ln x}{x})'=\frac{1-ln x}{x^2}$. When x>e, $\frac{1-ln x}{x^2}<0$, meaning the function $\frac{ln x}{x}$ is monotonically decreasing at that point. If e < n < n+1, then

$$\frac{ln a}{a}>\frac{ln b}{b}<\Rightarrow \frac{ln a}{ln b}>\frac{a}{b}$$

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.