Question

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

Answer 1

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

Answer 2

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

Answer 3

seems okay to me

Answer 4

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 \(x-1<c<x\)

then \[\frac{1}{1+c}>\frac{1}{x+1}\]

thus \[\ln\left(1+\frac{1}{x}\right)> \frac{1}{x+1}\]

Answer 5

Thanks guy! My partner had the second one, but I wasn't sure that's why I asked! Thanks a lot :)