Show that if x is positive, then ln(1+1/x)>(1/1+x)

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Show that if x is positive, then ln(1+1/x)>(1/1+x)

Mathematics
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i think this will work. start with \[\ln(1+\frac{1}{x})-\frac{1}{x+1}\] take the derivative and get \[-\frac{1}{x(x+1)^2}\] which is negative so long as x is positive, making your function decreasing for all x > 0
then note that \[\lim_{x\rightarrow \infty} \ln(1+\frac{1}{x})-\frac{1}{x+1}=0\] so you have a strictly decreasing function that goes to zero, meaning it must be positive
seems okay to me

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you could also use the MVT on the function \[f(x)=\ln(x+1)\] with a=x-1, b=x \[\ln\left(1+\frac{1}{x}\right)=\ln\left(\frac{x+1}{x}\right)=\ln(x+1)-\ln(x)\] \[=\ln(x+1)-\ln((x-1)+1)=\frac{\ln(x+1)-\ln((x-1)+1)}{x-(x-1)}=f'(c)=\frac{1}{1+c}\] where \(x-1\frac{1}{x+1}\] thus \[\ln\left(1+\frac{1}{x}\right)> \frac{1}{x+1}\]
Thanks guy! My partner had the second one, but I wasn't sure that's why I asked! Thanks a lot :)

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