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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4Possible duplicate of How to prove $(1+1/x)^x$ is increasing when $x>0$? – YuiTo Cheng Jun 26 '19 at 06:06
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For $x\in \mathbb{N}$ this result can be derived using the AM-GM inequality: $$(\frac{n.\frac{n+1}{n}+1}{n+1})\geq((\frac{n+1}{n})^n.(1)^{1})^{(\frac{1}{n+1})}$$ Using A.M. G.M. inequality(i.e A.M$\geq$G.M.) $$\Rightarrow\frac{(n+1)+1}{n+1}\geq(\frac{n+1}{n})^{\frac{n}{n+1}}$$ $$\Rightarrow(1+\frac{1}{n+1})^{n+1}\geq(1+\frac{1}{n})^{n}$$
Maximilian Janisch
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Abhra Abir Kundu
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2I 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 08 '13 at 09:53
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3@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 08 '13 at 10:00
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$$f(x) = \left ( 1+ \frac{1}{x} \right )^x$$
$$\log{f(x)} = x \log{\left( 1+ \frac{1}{x} \right )}$$
$$\frac{d}{dx}\log{f(x)} = \log{\left( 1+ \frac{1}{x} \right )} + \frac{x}{1+\frac{1}{x}}\left (\frac{-1}{x^2} \right ) = \log{\left( 1+ \frac{1}{x} \right )} - \frac{1/x}{1+ \frac{1}{x}}$$
or
$$\frac{d}{dx}\log{f(x)} = \log{\left( 1+ \frac{1}{x} \right )} + \frac{1}{1+ \frac{1}{x}} - 1 $$
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.
Ron Gordon
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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 08 '13 at 09:41
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@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 08 '13 at 10:04
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$\frac{d}{dx}\log(f(x)) = \frac{1}{f(x)},\frac{df(x)}{dx} $. Since $f(x) >0$ so too is $\frac{df(x)}{dx}$ – Michael Levy Nov 21 '17 at 14:22
