TITLE: An inequality $\frac1{(n+1)^{1/(n+1)}}-\frac1{n^{1/n}}\le \frac1{n+1}$
QUESTION [5 upvotes]: I have graphed the functions $f,g:\mathbb{R^+}\to\mathbb{R}$ defined by $$f(x)=\frac1{(x+1)^{1/(x+1)}}-\frac1{x^{1/x}}\mbox{ and } g(x)=\frac1{x+1}$$
and it seems like $f(x)\le g(x)$ for all $x>0$. However, I just need the following result... 

$$\frac1{(n+1)^{1/(n+1)}}-\frac1{n^{1/n}}\le \frac{1}{n+1}\ \ \ \  
\forall n\in\mathbb{Z}^+. $$

Would someone please point me in the right direction?

REPLY [6 votes]: Consider the function $f(x)=x^{-1/x}$ with derivative $f'(x) = f(x)\cdot \frac{\log x - 1}{x^2}$. You want to show that
$$
f(n+1) - f(n) \le \frac{1}{n+1}
$$
for all integer $n$. For $x \ge 1$, clearly $f(x) \le 1$. Then
$$
f(n+1) - f(n) = \int_n^{n+1} f'(x) dx \le \int_n^{n+1}\frac{\log x - 1}{x^2} dx
$$ 
The integrand is decreasing  for $x > e^{3/2} \approx 4.4$ and therefore 
$$
f(n+1) - f(n) \le \frac{\log n - 1}{n^2}
$$
for $n \ge 5$. The right hand side is easily seen to be $\le \frac{1}{n+1}$. For $n < 5$ the desired inequality can be checked directly.