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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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    $\begingroup$ Have you tried the obvious avenue of looking at the sign of the derivative? $\endgroup$ Nov 17, 2011 at 13:08
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    $\begingroup$ Perhaps by showing that log(F(x)) is increasing for x>0? $\endgroup$
    – ofer
    Nov 17, 2011 at 13:09
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    $\begingroup$ Pursuing ofer's suggestion, you'll need to use the inequality $\ln(1+x)\ge x$ for $x\ge0$. $\endgroup$ Nov 17, 2011 at 13:13
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    $\begingroup$ Combine the ideas of Henning Makholm and ofer, and look at the sign of the derivative of $\log(F(x))$. $\endgroup$
    – J126
    Nov 17, 2011 at 13:14
  • $\begingroup$ @David: Double-check the direction of that "inequality". (I'm sure it was just a typo.) :) $\endgroup$
    – cardinal
    Nov 17, 2011 at 13:26

8 Answers 8

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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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    $\begingroup$ 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 $x \to \infty$). $\endgroup$
    – Mike F
    Nov 18, 2011 at 8:31
  • $\begingroup$ And, of course, this shows the solution for all real positive $x$, since the rationals are dense in $\mathbb R$ and our function is smooth. $\endgroup$
    – YiFan Tey
    Feb 26, 2019 at 6:09
  • $\begingroup$ $b=1/(2n)$ is the correct choice to prove the boundedness by $4$. $\endgroup$
    – A.Γ.
    Sep 27, 2020 at 15:37
  • $\begingroup$ @A.Γ. thanks for catching my mistake. I think there's also a mistake in your correction, though. Did you mean to say $b=1+1/(2n)$ is the correct choice? $\endgroup$
    – Mike F
    Sep 27, 2020 at 22:25
  • $\begingroup$ Yea, right. "1+" was cut off. Finally, two wrong is one right =) $\endgroup$
    – A.Γ.
    Sep 28, 2020 at 9:15
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It's enough to show $\ln(1+1/x)^x = x\ln(1+1/x)$ is increasing. Letting $h=1/x$, we can write the last expression as

$$\tag 1 \frac{\ln(1+h)-\ln 1}{h}.$$

We want to show $(1)$ increases as $h$ decreases. Now $(1)$ is just the slope of the line through $(1,\ln 1)$ and $(1+h,\ln(1+h)).$ And any concave function has the property that such slopes increase as $h$ decreases. Since $\ln x$ is concave, we're done.

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  • $\begingroup$ Very elegant! $(+1)$ $\endgroup$
    – Joe
    Jun 29, 2021 at 18:29
  • $\begingroup$ @Joe Thank you. $\endgroup$
    – zhw.
    Jun 29, 2021 at 19:02
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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.

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  • $\begingroup$ Please, where do you get this inequality from $ h(x) = \mathrm e^{-1/(1+x)} > \left(1-\frac{1}{1+x}\right) $ ? $\endgroup$ Oct 3, 2019 at 21:11
  • $\begingroup$ Related math.stackexchange.com/a/3379553/198592, Corollary. $\endgroup$ Oct 3, 2019 at 21:17
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    $\begingroup$ Here's a proof of the relation I just asked you for: Starting from the obvious relation $1-\exp (-t)>0$ for $t>0$ the following integral $\int_0^{\frac{1}{x+1}} (1-\exp (-t)) \, dt =e^{-\frac{1}{x+1}}+\frac{1}{x+1}-1$ is also positive Q.E.D. $\endgroup$ Oct 3, 2019 at 21:46
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$(1+1/x)^x$ is increasing equivalent to $x\cdot\ln(1+1/x)$ is increasing by taking natural logarithm. Which is equivalent to derivative being positive, which is showing that $h(x) = \ln(1+1/x) - 1/(1+x) > 0$.

And now notice $h'(x)=-1/[x\cdot(x+1)^2]$ is negative, so $h(x)$ is decreasing and minimal value is at $+\infty$, but $h(+\infty)=0$. So, $h(x) \ge 0$.

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  • $\begingroup$ Your $h(x)$ is not the derivative of $x \log(x+1/x)$ or am I missing something? $\endgroup$ Jun 29, 2021 at 20:00
  • $\begingroup$ @principal-ideal-domain correct...it is the derivative of $x\log(1+\frac{1}{x})$, which is what he needs to determine. Not sure why you are talking about the function $x\log(x+1/x)$ $\endgroup$
    – S.C.
    Oct 17, 2023 at 23:57
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We have to prove that

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

is a strictly increasing function for $x>0$.

The derivative is

$$F'(x) = \frac{\left(\frac{1}{x}+1\right)^x \left((x+1) \log \left(\frac{1}{x}+1\right)-1\right)}{x+1}$$

Hence we only need to show that

$$\log \left(\frac{1}{x}+1\right)-\frac{1}{x+1}>0 $$

But this is obvious since the l.h.s. is equal to the definitely positive integral

$$\int_x^{\infty } \frac{1}{t (t+1)^2} \, dt$$

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  • $\begingroup$ What is the motivation to have seen this integral? It would have never crossed my mind to consider comparing to an integral, and I still don't see how one would produce the integrand (just checking it required paper for me). $\endgroup$ Nov 27, 2019 at 21:23
  • $\begingroup$ @ Trevor K The motivation was, of course, to find the proof I have given. And the integrand, well, I just "saw" it from looking at the inequality. BTW I find it's a pretty simple and elegant proof.. $\endgroup$ Nov 28, 2019 at 0:07
  • $\begingroup$ If you do say so yourself! Wow, I am astounded by your brilliance. If only I were as intelligent as you. $\endgroup$ Nov 28, 2019 at 0:09
  • $\begingroup$ @ Trevor K To the matter, please. $\endgroup$ Nov 28, 2019 at 0:13
  • $\begingroup$ Dear downvoter: please explain your decision. What is wrong with the proof? It would also be nice to know who you are. $\endgroup$ Nov 28, 2019 at 14:32
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This proof is from theorem 140 from Hardy's Inequalities. Let $f(x) = \ln\left[\left(1 + \frac{1}{x}\right)^x\right] = x(\ln(x+1) - \ln(x)).$ We refer to the mean value theorem: for each differentiable $g$, $$ g(x + h) - g(x) = hg'(x + \theta h) $$ for $\theta \in (0,1)$.

Applying the MVT to $g(x) = \ln(x)$, we get $\ln(x+1) - \ln(x) = \frac{1}{x + \theta}$. Since, $\frac{1}{x+\theta} > \frac{1}{x+1} $, we have that $\ln(x+1) - \ln(x) > \frac{1}{x+1}$, so $$ f'(x) = \ln(x+1) - \ln(x) - \frac{1}{x+1} > 0. $$ Hence, $f(x)$ and $e^{f(x)} = \left(1+\frac{1}{x}\right)^x$ are increasing with $x$.

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    $\begingroup$ How did you arrive at f′(b)=ln(x+1)−ln(x)+1x>0 ? $\endgroup$ Oct 25, 2020 at 8:35
  • $\begingroup$ It was a typo, sorry for the confusion. It is meant to say $f'(x) = ln(x+1) - ln(x) + \frac{1}{x}$. $\endgroup$
    – DSM
    Oct 25, 2020 at 17:31
  • $\begingroup$ How is ln(x+1)−ln(x)+1/x larger than zero? $\endgroup$ Oct 26, 2020 at 11:37
  • $\begingroup$ It was yet another mistake. I'm very sorry that there are so many! I had the wrong derivative and also used the wrong inequality. It should be correct now. $\endgroup$
    – DSM
    Oct 27, 2020 at 0:19
  • $\begingroup$ Ah that's an interesting way of proving it. $\endgroup$ Oct 27, 2020 at 21:30
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$f(t):=\frac{1}{t}\log(1+t)$ 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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    $\begingroup$ 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. $\endgroup$
    – cardinal
    Nov 17, 2011 at 13:40
  • $\begingroup$ @cardinal thanks for your careful reading, I hope to have fixed all my mistakes. $\endgroup$
    – agt
    Nov 17, 2011 at 13:49
  • $\begingroup$ I believe the derivative of $g(t)$ is still not quite right. :) (I believe, the whole denominator should be squared, i.e., $g'(t) = t/(1+t)^2$.) $\endgroup$
    – cardinal
    Nov 17, 2011 at 13:54
  • $\begingroup$ I have made some corrections. I hope you don't mind. $\endgroup$
    – cardinal
    Nov 17, 2011 at 17:17
  • $\begingroup$ @cardinal Thank you very much for your correction. I was in a hurry today. Excuse me. $\endgroup$
    – agt
    Nov 17, 2011 at 19:14
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The given inequality is equivalent to $$ \ln (x+1)-\ln(x) > \frac{1}{x+1},\ \ \ \ \ \ \ \ \ \ x\gt 0.$$ Let $f(t)= \frac{1}{t}$. Then, $$ \ln (x+1)-\ln(x) = \int_1^{x+1} f(t)\ dt - \int_1^x f(t)\ dt $$$$ \ \ \ \ \ =\int_x^{x+1} f(t)\ dt $$$$\ \ \ \ \ge 1\cdot f(x+1) $$$$\ \ \ \ \ \ \ \ \ \ \ \ \ =f(x+1) = \frac{1}{x+1}$$ since $f$ is a decreasing function.

The strict inequality can be obtained with a little more work; by breaking up the interval $[x, x+1]$ into two pieces - such as $\left[x, x+{1\over 2}\right]$ and $\left[x+{1\over 2}, x+1\right]$ - and applying the same argument as above on each interval.

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