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?
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.