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How does one prove that $(1+\frac{1}{x})^x$ is monotonic increasing for any $x \in [1,\infty)$?

Thanks a million!

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Duplicate. See… . In fact, it is proved for $x>0$. – vesszabo Nov 9 '15 at 19:42

For $x\in \mathbb{N}$ this result is very obvious. $$(\frac{n.\frac{n+1}{n}+1}{n+1})\geq((\frac{n+1}{n})^n.(1)^{1})^{(\frac{1}{n+1})}$$ Using A.M. G.M. inequality(i.e A.M$\geq$G.M.) $$\Rightarrow\frac{(n+1)+1}{n+1}\geq(\frac{n+1}{n})^{\frac{n}{n+1}}$$ $$\Rightarrow(1+\frac{1}{n+1})^{n+1}\geq(1+\frac{1}{n})^{n}$$

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I think he is learning the exponential function, at which point he may or may not have learnt derivatives, which is why I think this proof is better. – mez Feb 8 '13 at 9:53
@mezhang: I agree. Even if it were not so, isn't computing the derivative sort of cheating? I mean, the derivative of $\log(x)$ comes from the fact that $\lim_{x\to 0}(x+\frac 1x)^x=e$ and to show that such a limit exists you use the fact that the function is bounded and monotonous (considering $x\in\mathbb N$ is enough for the purpose). – AndreasT Feb 8 '13 at 10:00

$$f(x) = \left ( 1+ \frac{1}{x} \right )^x$$

$$\log{f(x)} = x \log{\left( 1+ \frac{1}{x} \right )}$$

$$\frac{d}{dx}\log{f(x)} = \log{\left( 1+ \frac{1}{x} \right )} + \frac{x}{1+\frac{1}{x}}\left (\frac{-1}{x^2} \right ) = \log{\left( 1+ \frac{1}{x} \right )} - \frac{1/x}{1+ \frac{1}{x}}$$


$$\frac{d}{dx}\log{f(x)} = \log{\left( 1+ \frac{1}{x} \right )} + \frac{1}{1+ \frac{1}{x}} - 1 $$

Because $y \log{y} \ge y-1 \; \forall y \ge 1$, we may say that, $\forall x \ge 1$:

$$\frac{d}{dx}\log{f(x)} =\log{\left( 1+ \frac{1}{x} \right )} + \frac{1}{1+ \frac{1}{x}} - 1 \ge 0$$

and $f(x)$ is monotone increasing for all $x \ge 1$, as $\log$ is monotone increasing.

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You missed a $x$ in the denominator of the second term in the last equality when computing the derivative. It should be $-\frac{1}{x\left(1 + \frac{1}{x}\right)}$. – Paresh Feb 8 '13 at 9:41
@Paresh: thank you, that was a major goof, not helped by the correct inequality. I have fixed the error and implemented the correct inequality. – Ron Gordon Feb 8 '13 at 10:04

Hint: Consider the sign of the derivative!

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