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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!

P. Gillich
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1 Answers1

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You need just to know that

$$\left(1+\frac{1}{x} \right)^{x+1}\ge e$$

then

$$\log{\left(1+\frac{1}{x} \right)}-\frac{1}{1+x}\ge{0} \iff \frac{\log{\left(1+\frac{1}{x} \right)}}{\frac{1}{1+x} }\ge{1}\iff \frac{1+x}{1+x}\frac{\log{\left(1+\frac{1}{x} \right)}}{\frac{1}{1+x} }\ge{1}\iff \frac{\log{\left(1+\frac{1}{x} \right)^{1+x}}}{\frac{1+x}{1+x} }\ge \log e={1}$$

user
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