# $n^n<(n-1)^{n+1}$ for integer $n\ge5$

For natural number $$n\ge5$$, by mathematical induction or otherwise, prove that $$n^n<(n-1)^{n+1}$$.

Actually I was trying to solve the problem that I posted.

## My Attempt:

Step I: Verify that when $$n=5$$, then the given inequality holds true;

$$5^5\overset{?}{<}4^6\Rightarrow 3125<4096$$ which is true.

Step II: Assume that the inequality is true for $$n=k\ge 5$$;

that is $$k^k<(k-1)^{k+1}$$ is true for $$k\ge5$$.

Step III: We need to prove that if the inequality is true for $$n=k$$, then it is true for $$n=k+1$$ as well;

that is to show that if $$k^k<(k-1)^{k+1}$$ is true, then $$(k+1)^{k+1} is also true.

I have a difficulty to complete this part of the proof. Any help would be appreciated.

I'd write $$m=n-1$$. Then the statement reduces to $$\left(1+\frac1m\right)^m<\frac{m^2}{m+1}.$$ It's well-known that $$(1+1/m)^m$$ increases to $$e$$, but more naively, $$\left(1+\frac1m\right)^m=1+1+\frac1{m^2}{m\choose 2} +\frac1{m^3}{m\choose 3}+\cdots<1+1+\frac12+\frac16+\cdots<3.$$ But $$\frac{m^2}{m+1}>\frac{m^2-1}{m+1}=m-1>3$$ if $$n>4$$.

• $m-1>3\iff m>4\iff n>5$, which means that you need to treat the case $n=5$ separately. Or use that because $e<3$ already is a strict inequality, and you only need one strict inequality in a chain to get strict inequality from first to last term, it is sufficient to demand $m-1\ge 3\iff n\ge 5$. – Lutz Lehmann Jul 9 '19 at 6:55

Here is a proof without induction.

$$(1-1/n)^n$$ is strictly increasing for $$n > 1$$ and approaches $$1/e$$ as $$n \to \infty$$ whereas $$1/(n-1)$$ is strictly decreasing and approaches $$0$$ as $$n \to \infty$$. Hence there is an $$N_0$$ such that for all $$n > N_0$$,

$$\Big(1 - \frac{1}{n}\Big)^n > \frac{1}{n-1}$$

By little computation, we find that smallest $$n$$ for which this inequality holds is $$n = 5$$. Since LHS is increasing and RHS is decreasing, it implies that the inequality will never be violated for $$n \ge 5$$.

The claim $$x^x<(x-1)^{x+1}$$ is equivalent to $$x\ln x<(x+1)\ln(x-1).$$ Consider $$f(x)=(x+1)\ln(x-1)-x\ln x$$ and prove that

1. $$f'(x)=\ln\left(1-\frac1x\right)+\frac2{x-1}\to 0$$ as $$x\to +\infty$$,
2. $$f''(x)=-\frac{x+1}{x(x-1)^2}<0$$ for $$x>1$$.

Then $$f'(x)>0$$ and $$f$$ increases. From $$f(5)>0$$ it follows that $$f(x)>0$$ for all $$x\ge 5$$.

• Or you can rewrite it to $\frac{\ln x}{x+1} < \frac{\ln (x-1)}x$ and show that the function $g(x) = \frac{\ln x}{x+1}$ is decreasing. – Quang Hoang Jul 9 '19 at 18:43
• @QuangHoang Yes, it is smart rewriting. – A.Γ. Jul 9 '19 at 19:04