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 \(\displaystyle{x}{>}{0}\)

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Expert Answer

Shawn Kim
Answered 2022-01-01 Author has 4997 answers
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 \(\displaystyle{t}{>}−{1}\),
\(\displaystyle{\log{{\left({1}+{t}\right)}}}{ < }{t}.\)
Your inequality is equivalent to
\(\displaystyle{\log{{\left({x}+{1}\right)}}}−{\log{{\left({x}\right)}}}={\log{{\left({1}+{\frac{{{1}}}{{{x}}}}\right)}}}{ < }{\frac{{{1}}}{{{x}}}}\)
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Edward Patten
Answered 2022-01-02 Author has 1439 answers
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.
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Vasquez
Answered 2022-01-09 Author has 8850 answers

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
\(x<c<x+1\Leftrightarrow\frac{1}{x+1}<\frac{1}{c}<\frac{1}{x}\)
\(\Rightarrow \frac{1}{x+1}<\ln(x+1)−\ln(x)<\frac{1}{x}\)

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