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Let $F(x)=(1+\frac{1}{x})^x$.

How do we prove $F(x)$ is increasing when $x>0$?

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    Have you tried the obvious avenue of looking at the sign of the derivative?2011-11-17
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    Perhaps by showing that log(F(x)) is increasing for x>0?2011-11-17
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    Pursuing ofer's suggestion, you'll need to use the inequality $\ln(1+x)\ge x$ for $x\ge0$.2011-11-17
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    Combine the ideas of Henning Makholm and ofer, and look at the sign of the derivative of $\log(F(x))$.2011-11-17
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    @David: Double-check the direction of that "inequality". (I'm sure it was just a typo.) :)2011-11-17
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    oops, thanks...2011-11-17
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    @Meredith: Is this homework? If so, please read the [FAQ](http://meta.math.stackexchange.com/questions/1803/how-to-ask-a-homework-question) regarding homework questions and update your post accordingly.2011-11-17

3 Answers 3

5

Let $h(x) = \mathrm e^{-1/(1+x)}(1+1/x)$ and note that for $x > 0$, $$ h(x) = \mathrm e^{-1/(1+x)}\cdot\frac{x+1}{x} > \left(1-\frac{1}{1+x}\right)\frac{x+1}{x} = 1. $$

Now, let $g(x) = \log F(x)$ and note that $$ g'(x) = \log(1+1/x) - \frac{1}{1+x} = \log h(x) > \log 1 = 0\>, $$ and so we are done.

5

There's an elementary approach for rational $x$. It suffices to prove that $$\left( 1+\frac{m}{n} \right)^n < \left( 1+\frac{m}{n+1} \right)^{n+1}$$ for $m,n$ positive integers. Whenever $0 \leq a < b$, we have $\frac{b^{n+1} - a^{n+1}}{b-a} = \sum_{k=0}^n a^{n-k}b^k < (n+1)b^n$ which rearranges to $$[(n+1)a - nb] \cdot b^n < a^{n+1}.$$

Substituting $a = 1+m/(n+1)$ and $b = 1 +m/n$ into the above, the term in square braces (miraculously) reduces to $1$ and we get the desired bound. This is adapted from Foundations of Mathematical Analysis by Johnsonbaugh and Pfaffenberger.

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    Incidentally, substituting $a=1$ and $b=1+2/n$, another miracle occurs and the term in square braces reduces to $1/2$. From this, one can deduce that $(1+1/x)^x$ is bounded by $4$ (hence converges as $x \to \infty$).2011-11-18
1

$f(t):=\frac{1}{t}\log(1+t)$ is decreasing for $t>0$, because it is smooth and its derivative is negative.

Its derivative is $f'(t)=-\frac{1}{t^2}g(t)$, where $g(t):=\log(1+t)-\frac{t}{1+t}.$
But, $g(t)>0$ for $t>0$, in fact, $\lim_{t\to 0}\ g(t)=0$ and $g$ is increasing.
$g$ is increasing because it is smooth and its derivative is positive: $g'(t)=\frac{t}{(1+t)^2}$.

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    Please correct this answer. $f(t)$ is *not* increasing in $t$ and $g'(t)$ is *not* negative, nor does it have the form you provide. The proof can be patched by considering these things carefully.2011-11-17
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    @cardinal thanks for your careful reading, I hope to have fixed all my mistakes.2011-11-17
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    I believe the derivative of $g(t)$ is still not quite right. :) (I believe, the whole denominator should be squared, i.e., $g'(t) = t/(1+t)^2$.)2011-11-17
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    I have made some corrections. I hope you don't mind.2011-11-17
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    @cardinal Thank you very much for your correction. I was in a hurry today. Excuse me.2011-11-17
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    No apology necessary. Cheers. :)2011-11-18
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    Can we also prove this using similar technique? https://math.stackexchange.com/questions/2287547/prove-that-1-frac-logxxx-is-monotonically-increasing-function-for-x2017-05-19