Is $\ln(1+\frac{1}{x-1}) \ge \frac{1}{x}$ for all $x \ge 2$? Plotting both functions  $\ln(1+\frac{1}{x-1})$ and $\frac{1}{x}$ in $[2,\infty)$ gives the impression that $\ln(1+\frac{1}{x-1}) \ge \frac{1}{x}$ for all $x \ge 2$.
Is it possible to prove it?
 A: Since $f(t)=\frac{1}{t}$ is a decreasing function over $\mathbb{R}^+$, for any $u>1$ we have:
$$ \log\left(1+\frac{1}{u-1}\right)=\log(u)-\log(u-1)=\int_{u-1}^{u}\frac{dt}{t} \geq \frac{1}{u}.\tag{1}$$
A: We know that for $|x| < 1$, $$\log(1-x) = -x-\frac{1}{2}x^2 - \frac{1}{3}x^3 - \dotsb.$$
Since $$1 + \frac{1}{x-1} = \frac{x-1+1}{x-1} = \frac{x}{x-1},$$
we see that 
$$\log \left(1+\frac{1}{x-1}\right) = \log \left(\frac{x}{x-1}\right) = -\log\left(\frac{x-1}{x}\right) = -\log\left(1-\frac{1}{x}\right).$$
So if $x \ge 2$ then $0 < \frac{1}{x} < 1$, so we can apply the power series:
$$-\log\left(1-\frac{1}{x}\right) = \frac{1}{x} + \frac{1}{2x^2} + \frac{1}{3x^3} + \dotsb > \frac{1}{x}.$$
A: Expand $\ln(1+\frac{1}{x-1})=\ln(\frac{x}{x-1})$ using Taylor series. You will get:
$$\ln(\frac{x}{x-1})=\frac{1}{x-1}-\frac12\left(\frac{1}{x-1}\right)^2+O(x^3)$$
Now you have to show that:
$$\frac{1}{x-1}-\frac12\left(\frac{1}{x-1}\right)^2\geq\frac1x$$
$$\frac{2x-3}{2\cdot(x-1)^2}\geq\frac1x$$
Which is equivalent to:
$$x>1$$
A: hint :
$$f(x)=ln(1+\frac{1}{x-1})-\frac{1}{x}\\x≥2\\ f'<0$$f(x) is decreasing function ,but f(x) is above the x axis $$\\f(2)>0,f(\infty)>0\\f(x)>0\\so\\ln(1+\frac{1}{x-1})-\frac{1}{x}≥0\\ln(1+\frac{1}{x-1})≥\frac{1}{x}$$
