$\ln( n + 1 ) - \ln( n ) > \frac 1{n+1}$
Is this statement true? I tried to show by $$\ln( n+1 /n)\implies 1+ 1/n > 0, \quad n >1$$ That is all I could get to so...
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2 Answers
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First Method
Let $f(x)=\ln(x+1)-\ln x-\frac{1}{1+x}$. $$f'(x)=\frac{1}{x+1}-\frac{1}{x}+\frac{1}{(1+x)^2}=\frac{x(x+1)-(x+1)^2+x}{x(x+1)^2}=...<0$$ for all $x>0$ and thus $f$ is strictly decreasing on $]0,+\infty [$. Moreover, $$\lim_{x\to 0^+ }f(x)=+\infty $$ and $$\lim_{x\to+\infty }f(x)=0$$ Therefore, $f(x)>0$ for all $x>0$.
Second Method
$$(n+1)\left(\ln(n+1)-\ln n\right)=\ln\left(\left(\frac{n+1}{n}\right)^{n+1}\right)=\ln\left(\left(1+\frac{1}{n}\right)^{n}\left(1+\frac{1}{n}\right)\right).$$
You have that $$\lim_{n\to\infty }\ln\left(\left(1+\frac{1}{n}\right)^{n}\left(1+\frac{1}{n}\right)\right)=\ln(e)=1.$$
Moreover, $x\mapsto \ln x$ is increasing and thus, if you set $$x_n=(n+1)\left(\ln(n+1)-\ln(n)\right)=\ln\left(\left(1+\frac{1}{n}\right)^{n}\left(1+\frac{1}{n}\right)\right),$$ $(x_n)$ is decreasing. Indeed, it's easy to show that $$y_n=\left(1+\frac{1}{n}\right)^{n+1}$$ is decreasing, and thus that $x_n=\ln(y_n)$ is also decreasing. Therefore $x_n>1$ for all $n$ and thus $$(n+1)\left(\ln(n+1)-\ln n\right)>1\implies \ln(n+1)-\ln n>\frac{1}{n+1}$$ what conclude the proof.
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$\begingroup$ Nice method of approach , do you also know other methods?because I like to see on different angles $\endgroup$– SocreCommented Mar 14, 2015 at 18:00
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$\begingroup$ I wrote an other method and my previous answer. $\endgroup$– SurbCommented Mar 14, 2015 at 18:31
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$\begingroup$ Excellent method if I had the " rep " to vote I would have! $\endgroup$– SocreCommented Mar 14, 2015 at 18:37
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$\begingroup$ no problem, I'm happy to have helped you :-) $\endgroup$– SurbCommented Mar 19, 2015 at 1:30
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here is another way to look at this $$\ln(n + 1) -\ln n = \ln \frac{n+1}{n} = \ln \left( 1 + \frac 1 n\right) = \int_1^{1 + \frac 1n} \frac{dx}{x} < \frac 1 n \frac 1{\left(1 + \frac 1 n\right)} = \frac 1{n+1}$$
the reason for the inequality is that the function $\frac 1 x$ is decreasing on the interval $1 \le x \le 1 + \frac 1n.$
