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I am struggling to see how to use the following two pieces of information (which I have proven):

$\frac{1}{n}< \ln{n}$ for $n=2,3,4,\dots$

and

$\ln{(1+n)}<n$ for $n=1,2,3,\dots$

to show that

$\frac{n}{1+n}<\ln{\left(1+\frac{1}{n}\right)^n}<1$ for $n=1,2,3,\dots$

I know it's more than likely something very straight forward.

Did
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RedG
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  • Do you need to use both inequalities and can you use the properties of logarithms?Also is it $(\ln(1+\frac{1}{n}))^n$ or $\ln(1+\frac{1}{n})^n$ I assume it's the second. – kingW3 Mar 19 '17 at 15:42
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    To be proven: $$\frac1{n+1}<\ln(1+x)<\frac1n\qquad x=\frac1n$$ The rightmost inequality is direct from $\ln(1+u)<u$, valid for every $u>-1$, $u\ne0$. The leftmost inequality is direct from $$\ln (1+x)=-\ln(1-y)\qquad y=\frac1{n+1}$$ using once again the fact that $\ln(1+u)<u$. (One can see that the (rather bizarre) first "piece of information" in your question is not useful here. It rarely is.) – Did Mar 19 '17 at 15:44

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