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

1. anonymous

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

2. anonymous

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

3. anonymous

seems okay to me

4. Zarkon

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 $\ln\left(1+\frac{1}{x}\right)> \frac{1}{x+1}$

5. anonymous

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