Prove $n^{n+1}$ is greater than $(n+1)^n$ for all $n > 2$

Question

The question of whether $2014^{2015}$ or $2015^{2014}$ is greater came up in my Calculus class and it seems clear that $2014^{2015} > 2015^{2014}$ based on a table comparing the sequences $s_n = n^{n+1}$ and $t_n = (n+1)^n$ for $n > 2$. However, I am unable to prove this result.

Thus far, I attempted an inductive proof without success as well as recasting the question as an analysis question where $f(x) = x^{x+1}$ and $g(x) = (x+1)^x$. It is easy to show that $f(3) > g(3)$, thus, if it could be shown that $f'(x) > g'(x)$ for all $x \geq 3$ the proof would follow easily. However, the derivatives are not `nice.'

Any help would be appreciated!

Answer 1

$n^{n+1}/(n+1)^n = n ( n^n / (n+1)^n) = n ( 1/(1+1/n)^n) \geq n * 1/e >1 $ for $n>2$ since $(1+1/n)^n \to e$ monotonically increasing. Thus, $n^{n+1} > (n+1)^n$.

Answer 2

$$n^{n+1}>(n+1)^{n}\Leftrightarrow(n+1)\ln n>n\ln(n+1)\Leftrightarrow\frac{\ln n}{n}>\frac{\ln(n+1)}{n+1}$$ Set $f(x)=\frac{\ln x}{x}$, then $f'(x)=\frac{1-\ln x}{x^{2}}<0$ when $x>e$, so $f(n+1)<f(n)$ when $n\geq3$.

Answer 3

Beside Batman's answer, from a purely algebraic point of view, you also could show that, for large values of $x$ $$f(x)=\frac{x^{x+1}}{(x+1)^x}=\frac{x}{e}+\frac{1}{2 e}-\frac{5}{24 e x}+\frac{5}{48 e x^2}+O\left(\left(\frac{1}{x}\right)^3\right)$$ and, using the first and second terms, $f(x) >1$ as soon as $x>e-\frac{1}{2}\approx 2.21828$, while the rigorous solution is $2.29317$