Author: Scott Karren

117. Show that if [latex]x[/latex] is positive, then [latex]\log_e\left(1+\dfrac{1}{x}\right)>\dfrac{1}{1+x}[/latex] First, note that [latex]\begin{align*}\log_e\left(1+\dfrac{1}{x}\right)&=\ln\left(\frac{x}{x}+\frac{1}{x}\right)\\&=\ln\left(\dfrac{x+1}{x}\right)\\&=\ln(x+1)-\ln(x)\\&=\int_x^{1+x}\dfrac{du}{u}\end{align*}[/latex]   Note also that for all [latex]u[/latex] between [latex]x[/latex] and [latex]x+1[/latex], the inequality: [latex]x\le u\le x+1[/latex] implies…