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anonymous
 4 years ago
Show that if x is positive, then ln(1+1/x)>(1/1+x)
anonymous
 4 years ago
Show that if x is positive, then ln(1+1/x)>(1/1+x)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then 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

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.3you could also use the MVT on the function \[f(x)=\ln(x+1)\] with a=x1, b=x \[\ln\left(1+\frac{1}{x}\right)=\ln\left(\frac{x+1}{x}\right)=\ln(x+1)\ln(x)\] \[=\ln(x+1)\ln((x1)+1)=\frac{\ln(x+1)\ln((x1)+1)}{x(x1)}=f'(c)=\frac{1}{1+c}\] where \(x1<c<x\) then \[\frac{1}{1+c}>\frac{1}{x+1}\] thus \[\ln\left(1+\frac{1}{x}\right)> \frac{1}{x+1}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks guy! My partner had the second one, but I wasn't sure that's why I asked! Thanks a lot :)
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