I am trying to show that the function $f(x)=(1+\frac1x)^x$ on $[0,+\infty)$.

I have found that $f'(x)=f(x) \left[\ln\left(\frac{x+1}x\right)- \frac1{x+1}\right]$.

Since $f(x)$ is always positive, I only have to show that $$\ln\left(\frac{x+1}x\right)>\frac1{x+1}\text{ when }x>0.$$

Is there an easy way to do this?

  • 1
    $\begingroup$ You don't need the word "always". Increasing on $(0,\infty)$ says it all. $\endgroup$
    – zhw.
    Sep 25, 2017 at 23:57
  • 7
    $\begingroup$ This follows from the standard inequality $$\frac{x} {x+1}<\log(1+x)<x$$ for all $x>0 $ which is obtained by integrating the inequality $$\frac{1}{x+1}<\frac{1}{1+t}<1$$ with respect to $t$ on interval $[0,x]$. Your desired result follows from the first part of this inequality if you replace $x$ by $1/x$. $\endgroup$
    – Paramanand Singh
    Sep 26, 2017 at 4:03
  • $\begingroup$ The following more recent post is a multiduplicate: Prove that $\ln\left( \frac{1+x}x\right) >\frac{1}{1+x}$? $\endgroup$ Jun 10, 2023 at 13:56

7 Answers 7


A different approach altogether:

If we start with Bernoulli's inequality, $(1+u)^r\gt1+ru$ for $u\ge0$ and $r\ge1$ (which is easy to prove by taking the derivative of $f(u)=(1+u)^r-1-ru$), we have, on letting $u=rx$,

$$\left(1+{1\over xr}\right)^r\ge1+r\cdot{1\over rx}=1+{1\over x}\implies\left(1+{1\over rx}\right)^{rx}=\left(\left(1+{1\over rx}\right)^r\right)^x\ge\left(1+{1\over x}\right)^x\quad\text{if }r\ge1$$

hence if $y=rx\ge x$ then

$$\left(1+{1\over y}\right)^y\ge\left(1+{1\over x}\right)^x$$

  • $\begingroup$ Very nice+1. Bernoulli inequality never ceases to amaze me! $\endgroup$
    – Paramanand Singh
    Sep 26, 2017 at 4:05

$$ \begin{align} \log\left(1+\frac1x\right) &=\int_0^{1/x}\frac{\mathrm{d}t}{1+t}\\ &\ge\int_0^{1/x}\frac{\mathrm{d}t}{1+1/x}\\[3pt] &=\frac1{x+1} \end{align} $$

  • 2
    $\begingroup$ How did you get from the first line to the second line? $\endgroup$
    – 1123581321
    Sep 26, 2017 at 0:55
  • $\begingroup$ @1123581321 When in doubt, draw a picture. i.e. sketch $1/(1+t)$ between $0$ and $1/x$. Now draw $1/(1+1/x)$ over the same values of $t$. The required argument should be completely obvious to you as soon as you do. $\endgroup$
    – Glen_b
    Sep 26, 2017 at 2:16
  • 1
    $\begingroup$ $1/x \ge t$ for $0 \le t \le 1/x$. $\endgroup$ Sep 26, 2017 at 2:16
  • $\begingroup$ It is sort of amusing that I used this very inequality in my previous answer. $\endgroup$
    – robjohn
    Sep 26, 2017 at 13:16

$$\ln\frac{x+1}x>\frac1{x+1}$$ $$\iff-\ln\left(1-\frac1{x+1}\right)>\frac1{x+1}$$ For $x>0$, $0<\frac1{x+1}<1$, so substitute $y=\frac1{x+1}$: $$\iff-\ln(1-y)>y$$ The Maclaurin series of $\ln(1-y)$ is always valid for $0<y<1$: $$\iff y+\frac{y^2}2+\frac{y^2}3+\dots>y$$ $$\iff\frac{y^2}2+\frac{y^3}3+\dots>0$$ which is true since $y$ is positive.


Let $g(x) = \ln\left(\frac{x+1}x\right)-\frac1{x+1} = \ln(x+1)-\ln(x)-\frac1{x+1} $.

$\begin{array}\\ g'(x) &=\frac1{x+1}-\frac1{x}+\frac1{(x+1)^2}\\ &=\frac{x(x+1)-(x+1)^2+x}{x(x+1)^2}\\ &=\frac{-1}{x(x+1)^2}\\ \end{array} $

Therefore $g(x)$ is decreasing.

But $\lim_{x \to \infty} \ln(1+1/x) = 0 $ and $\lim_{x \to \infty} \frac1{1+x} = 0 $, so that $\lim_{x \to \infty} g(x) = 0 $.

Therefore $g(x)> 0$ for $x > 0$.


This is an old problem and there are already clever solutions (see robjhon's solution above using integration) and the comment by Paramanand Sing (using classical inequalities). Both strategies solve the problem under the assumption $x>0$, stated by the OP.

Here I just want to add that the statement is also valid for $x<-1$. The method follows the along the lines proposed by the OP, namely, to look at the sign of $f'(x)$. The whole method relies on a simple extension the inequalities mentioned by Paramanand Sing to all $x+1>0$.

Proposition: If $t+1>0$, then $$\frac{t}{1+t}\leq \log(1+t)\leq t$$ with equality iff $t=0$.

Proof: The right hand side follows immediately as $1+t\leq e^t$ for all $t\in\mathbb{R}$ with equality iff $t=0$. The left hand side can be obtained from $$\frac{1}{1+t}=1-\frac{t}{1+t}\leq \exp\big(-\frac{t}{1+t}\big)$$ by taking logarithms on both sides of this inequality.

Using the computation of $f'(x)=f(x)\Big(\log\big(1+\tfrac{1}{x}\big)-\frac{1}{1+x}\Big)$ already provided by the OP, we obtain that for $x<-1$ or $x>0$, $1+\frac{1}{x}>0$ and so, $$\log\big(1+\tfrac{1}{x}\big)-\frac{1}{1+x}=\log\big(1+\tfrac{1}{x}\big)-\frac{\tfrac1x}{1+\tfrac1x}>0$$

Putting things together, we obtain that $f$ is monotone increasing in $(-\infty,-1)$, and also monotone increasing in $(0,\infty)$.

Comment: The same technique works for showing that $g:x\mapsto \big(1+\frac{1}{x}\big)^{x+1}$ is monotone decreasing: for all $x<-1$ or $x>0$ $$g'(x)=g(x)\Big( \log(1+\tfrac1x)-\tfrac{1}{x}\Big)<0$$ which follows by the proposition above.

All these yield following useful inequality $$\Big(1+\frac{1}{x}\Big)^x<e<\Big(1+\frac{1}{x}\Big)^{x+1}$$ for all $x>0$ and $$\Big(1+\frac{1}{x}\Big)^{x+1}<e<\Big(1+\frac{1}{x}\Big)^x$$ for all $x<-1$.


$$x>0\implies\log (1+1/x)=\log (1+x)-\log x=$$ $$=\int_x^{x+1}(1/t)dt>\int_x^{x+1}(1/(x+1)dt=1/(x+1).$$


Theorem 1. For $x>0$, the function $f_{\alpha}(x)=\bigl(1+\frac1x\bigr)^{x+\alpha}$ increases if and only if $\alpha\le0$ and decreases if and only if $\alpha\ge\frac12$.

For $x<-1$, the function $f_{\alpha}(x)$ decreases if and only if $\alpha\ge1$ and increases if and only if $\alpha\le\frac{1}{2}$.

The necessary and sufficient conditions such that the sequence $a_n=\bigl(1+\frac1n\bigr)^{n+\alpha}$ decreases or increases are $\alpha\ge\frac12$ or $\alpha\le\frac{2\ln3-3\ln2}{2\ln2-\ln3}$ respectively.

Proof. Direct calculation gives \begin{align*} [\ln f_\alpha(x)]'&=\ln\Bigl(1+\frac1x\Bigr) -\frac{x+\alpha}{x(x+1)}& \text{and}& & [\ln f_\alpha(x)]''&=\frac{(2\alpha-1)x+\alpha}{x^2(x+1)^2}. \end{align*}

For $x>0$, it is easy to see that $[\ln f_\alpha(x)]''>0$ and $[\ln f_\alpha(x)]'$ increases if and only if $\alpha\ge\frac12$. Since $\lim_{x\to\infty}[\ln f_\alpha(x)]'=0$ for any $\alpha\in\mathbb{R}$, thus $[\ln f_\alpha(x)]'<0$ for $\alpha\ge\frac12$, which means $f_\alpha'(x)<0$ and $f_\alpha(x)$ decreases. This implies also that the sequence $a_n$ is decreasing for $\alpha\ge\frac12$.

For $x>0$, it is clear that $[\ln f_\alpha(x)]''<0$ and $[\ln f_\alpha(x)]'$ decreases if and only if $\alpha\le0$. Then $[\ln f_\alpha(x)]'>0$, $f_\alpha'(x)>0$ and $f_\alpha(x)$ increases for $\alpha\le0$. This implies that the sequence $\bigl(1+\frac1n\bigr)^{n+\alpha}$ is increasing for $\alpha\le0$.

For $x>0$, when $0<\alpha<\frac12$, the function $[\ln f_\alpha(x)]''$ has a unique zero point $x_0=\frac{\alpha}{1-2\alpha}>0$ which is a supremum point of $[\ln f_\alpha(x)]'$, this supremum equals $[\ln f_{\alpha}(x_0)]'=\ln\bigl(\frac1\alpha-1\bigr)+2(2\alpha-1)>0$. Since $\lim_{x\to0^+}[\ln f_\alpha(x)]'=-\infty$ for $\alpha>0$ and $\lim_{x\to\infty}[\ln f_\alpha(x)]'=0$ for any $\alpha\in\mathbb{R}$, it is yielded that the functions $[\ln f_\alpha(x)]'$ and $f_\alpha'(x)$ have only one zero point $x_1>0$, which is a unique infimum point of $f_\alpha(x)$ on $(0,\infty)$. Consequently, the sufficient and necessary condition of the sequence $a_n$ being increasing is $f_\alpha(1)\le f_\alpha(2)$ which is equivalent to $\alpha\le\frac{2\ln3-3\ln2}{2\ln2-\ln3}$.

For $x<-1$, the function $[\ln f_\alpha(x)]''>0$ and $[\ln f_\alpha(x)]'$ is increasing if and only if $\alpha\le\frac12$. From $\lim_{x\to-\infty}[\ln f_\alpha(x)]'=0$ it is deduced that $[\ln f_\alpha(x)]'>0$ and $f_\alpha'(x)>0$ in $(-\infty,-1)$. Consequently, the function $f_\alpha(x)$ is increasing in $(-\infty,-1)$ if $\alpha\le\frac12$.

For $x<-1$, the function $[\ln f_\alpha(x)]''<0$ and $[\ln f_\alpha(x)]'$ is decreasing if and only if $\alpha\ge1$. From $\lim_{x\to-\infty}[\ln f_\alpha(x)]'=0$ it follows that $[\ln f_\alpha(x)]'<0$ and $f_\alpha'(x)<0$ in $(-\infty,-1)$. Accordingly, the function $f_\alpha(x)$ decreases in $(-\infty,-1)$ if $\alpha\ge1$.

For $x<-1$ and $\frac12<\alpha<1$, the function $[\ln f_\alpha(x)]''$ has a unique zero point $x_0=\frac{\alpha}{1-2\alpha}<-1$ which is a minimum point of $[\ln f_\alpha(x)]'$. Since $\lim_{x\to(-1)^-}[\ln f_\alpha(x)]'=\infty$ and $\lim_{x\to-\infty}[\ln f_\alpha(x)]'=0$, then the functions $[\ln f_\alpha(x)]'$ and $f_\alpha'(x)$ have only one zero point $x_1>0$, which is a unique infimum point of $f_\alpha(x)$ on $(0,\infty)$. This completes the proof of Theorem 1.

Note. The above theorem and its proof are extracted from the paper [3]. For more information on further developments, please refer to the papers [1,2,4] and closely related references therein.


  1. C. Berg, E. Massa, and A. P. Peron, A family of entire functions connecting the Bessel function $J_1$ and the Lambert $W$ function, Constr. Approx. 53 (2021), no. 1, 121--154; available online at https://doi.org/10.1007/s00365-020-09499-x.
  2. Bai-Ni Guo and Feng Qi, A property of logarithmically absolutely monotonic functions and the logarithmically complete monotonicity of a power-exponential function, University Politehnica of Bucharest Scientific Bulletin Series A---Applied Mathematics and Physics 72 (2010), no. 2, 21--30.
  3. Feng Qi, Wei Li, and Bai-Ni Guo, Generalizations of a theorem of I. Schur, Applied Mathematics E-Notes 6 (2006), Article 29, 244--250.
  4. Feng Qi and Aying Wan, A closed-form expression of a remarkable sequence of polynomials originating from a family of entire functions connecting the Bessel and Lambert functions, Sao Paulo Journal of Mathematical Sciences 16 (2022), no. 2, 1238--1248; available online at https://doi.org/10.1007/s40863-021-00235-2.

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