 Show that \log(x+1)−\log(x)<\frac{1}{x} for x>0 Cynthia Bell 2021-12-31 Answered
Show that $$\displaystyle{\log{{\left({x}+{1}\right)}}}−{\log{{\left({x}\right)}}}{ < }{\frac{{{1}}}{{{x}}}}$$ for $$\displaystyle{x}{>}{0}$$

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it Shawn Kim
The function $$\displaystyle{\log{{\left({1}+{t}\right)}}}$$ is strictly concave and therefore its graph stays under its tangent line at $$\displaystyle{0}:$$ for any $$\displaystyle{t}\ne{0}$$ and $$\displaystyle{t}{>}−{1}$$,
$$\displaystyle{\log{{\left({1}+{t}\right)}}}{ < }{t}.$$
$$\displaystyle{\log{{\left({x}+{1}\right)}}}−{\log{{\left({x}\right)}}}={\log{{\left({1}+{\frac{{{1}}}{{{x}}}}\right)}}}{ < }{\frac{{{1}}}{{{x}}}}$$
Not exactly what you’re looking for? Edward Patten
Consider $$\displaystyle{f{{\left({x}\right)}}}={\log{{\left({\frac{{{x}+{1}}}{{{x}}}}\right)}}}={\log{{\left({1}+{\frac{{{1}}}{{{x}}}}\right)}}}$$ . Set $$\displaystyle{u}={\frac{{{1}}}{{{x}}}}$$ and then after making a good figure you will see that $$\displaystyle{\log{{\left({1}+{u}\right)}}}\leq{u}$$ and you are done. Vasquez

Let $$x>0$$.
$$f:t\rightarrow\ln(t)$$ is continuous at $$[x,x+1]$$ and differentiable at $$(x,x+1)$$, thus by MVT,
$$\exists c \in(x,x+1):$$
$$f(x+1)−f(x)=(x+1−x)f′(c)$$
$$\Rightarrow\ln(x+1)-\ln(x)=\frac{1}{c}$$
and
$$x<c<x+1\Leftrightarrow\frac{1}{x+1}<\frac{1}{c}<\frac{1}{x}$$
$$\Rightarrow \frac{1}{x+1}<\ln(x+1)−\ln(x)<\frac{1}{x}$$