I’ve been trying to solve this for hours and it’s driving me crazy; this problem arose when trying to show that the function
$$\left(1+\frac{1}{x} \right)^x$$
Is always increasing for x positive. I know it suffices to show that the log of this function’s derivative is positive on the same interval, however this leads to showing that:
$$\log{\left(1+\frac{1}{x} \right)}-\frac{1}{1+x}\ge{0}$$
For all x positive, and log is the natural logarithm
I managed to show this is true if x is greater than or equal to 1, but I’m unsure how to proceed to show it’s true for the rest of the interval.
Please help! It’s driving me crazy!
You need just to know that
$$\left(1+\frac{1}{x} \right)^{x+1}\ge e$$
then
$$\log{\left(1+\frac{1}{x} \right)}-\frac{1}{1+x}\ge{0} \iff \frac{\log{\left(1+\frac{1}{x} \right)}}{\frac{1}{1+x} }\ge{1}\iff \frac{1+x}{1+x}\frac{\log{\left(1+\frac{1}{x} \right)}}{\frac{1}{1+x} }\ge{1}\iff \frac{\log{\left(1+\frac{1}{x} \right)^{1+x}}}{\frac{1+x}{1+x} }\ge \log e={1}$$