# Ngô Quốc Anh

## March 31, 2021

### Monotonicity of (1+1/x)^(x+α)

Filed under: Uncategorized — Ngô Quốc Anh @ 1:11

This post concerns the monotonicity of $\displaystyle f: x \mapsto \Big(1+\frac 1x\Big)^{x+\alpha}$

on $(0,+\infty)$ with $\alpha \geq 0$. The two cases $\alpha=0$ and $\alpha=1$ are of special because these are always mentioned in many textbooks as $\displaystyle \lim_{n \to \infty} \Big(1+\frac 1n\Big)^n= \lim_{n \to \infty} \Big(1+\frac 1n\Big)^{n+1} =e.$

Clearly, $f$ is monotone increasing with respect to $\alpha$. Hence we are left with the monotonicity of $f$ with respect to $x$. To study this problem, we examine $f'$ with respect to $x$.

Derivative of $f$. It is easy to get $\displaystyle f'(x) = \Big(1+\frac 1x\Big)^{x+\alpha} \Big[ \log \Big(1+\frac 1x\Big)- \frac{x+a}{x(x+1)}\Big].$

Hence the sign of $f'$ is determined by the sign of $\displaystyle g(x) = \log \Big(1+\frac 1x\Big)- \frac{x+a}{x(x+1)} .$

on $(0, \infty)$.

Obviously, when $a=0$ we have $\displaystyle f'(x) = \Big(1+\frac 1x\Big)^{x+\alpha} \Big[ \log \Big(1+\frac 1x\Big)- \frac 1{x+1}\Big] \leq 0$

and when $a=1$ we have $\displaystyle f'(x) = \Big(1+\frac 1x\Big)^{x+\alpha} \Big[ \log \Big(1+\frac 1x\Big)- \frac 1{x}\Big] \geq 0.$

Consequently, we recover the monotonicity of $f$ in the two cases $a=0,1$.

Derivative of $g$. It is easy to get $\displaystyle g'(x) = \frac {(2a-1)x+a}{x^2(x+1)^2}.$

Hence the sign of $g'$ is determined by the sign of $(2a-1)x+a$

on $(0, \infty)$. Clearly we have three possible cases: $0 \leq a < 1/2$, $a=1/2$, and $a>1/2$. For the case $a \geq 1/2$, this is easy to handle because $(2a-1)x \geq 0$ which immediately implies $\displaystyle g'(x) = \frac {(2a-1)x+a}{x^2(x+1)^2} > 0$

everywhere on $(0, \infty)$. For the case $a < 1/2$, we note that $(2a-1)x+a \leq 0$ if $x \geq a/(1-2a)$. Hence putting the above two cases together we obtain \displaystyle g'(x) \left\{ \begin{aligned} > 0 & \quad \text{ if } a \geq 1/2\\ >0 & \quad \text{ if } a< 1/2, \quad 0

Once we know precisely the sign of $g'$, we can examine the sign of $g$. This part is a little bit sticky as we are dealing with both the log and fractional functions. Again, we separate the two cases $a \ne 1/2$ and $a=1/2$.

Estimate of $g(\infty)$ when $a \ne 1/2$. Via Taylor expansion up to two terms we obtain \displaystyle \begin{aligned} g(x) &= \frac 1x - \frac 1{2x^2} + o\Big( \frac 1{x^2} \Big) - \frac{x+a}{x(x+1)}\\ &=\frac{1-2a}{2x(x+1)} -\frac 1{2x^2(x+1)}+ o \Big( \frac 1{x^2} \Big) \\&= \frac{1-2a}2 \frac 1{x(x+1)} + o \Big( \frac 1{x^2} \Big). \end{aligned}

Hence for $x$ large enough, we easily get \displaystyle g(x) \left\{ \begin{aligned} < 0 & \quad \text{ if } a > 1/2\\ >0 & \quad \text{ if } a< 1/2.\end{aligned}\right.

It is crucial to realize that the coefficient $(1-2a)/2$ of the leading term does not vanish. Hence together with the monotonicity of $g$ we have just shown that

• if $a>1/2$, then $g$ is monotone increasing with $g(\infty) < 0$ and
• if $a<1/2$, then $g$ is monotone decreasing on $(\frac a{1-2a}, +\infty)$ with $g(\infty) < 0$.

Estimate of $g(\infty)$ when $a =1/2$. The above argument does not work since $g(x)= o(x^{-2})$. Hence, in this case, we need an extra work. Still by Taylor expansion up to three terms we obtain \displaystyle \begin{aligned} g(x) &= \frac 1x - \frac 1{2x^2} +\frac 1{3x^3}+ o\Big( \frac 1{x^3} \Big) - \frac{x+1/2}{x(x+1)}\\ &=-\frac 1{6x^2(x+1)} +\frac 1{3x^3(x+1)}+ o \Big( \frac 1{x^3} \Big) \\&=- \frac 16 \frac 1{x^2(x+1)} + o \Big( \frac 1{x^3} \Big). \end{aligned}

Hence for $x$ large enough, we easily get $g(x) < 0$.

Conclusion. From the above computation, we know that

• if $a \geq 1/2$, then $f$ is monotone decreasing on $(0, \infty)$ and
• if $0\leq a < 1/2$, then $f$ is monotone increasing on $(\frac a{1-2a}, \infty)$.

Application. Using the monotonicity in $\alpha$, we easily get $\displaystyle \Big( 1 + \frac 1n \Big)^n \leq \cdots \leq \Big( 1 + \frac 1n \Big)^{n+\frac 14} \leq \cdots \leq \Big( 1 + \frac 1n \Big)^{n+\frac 12} \leq \cdots \leq \Big (1 + \frac 1n \Big)^{n+1}.$

In addition, using the monotone decreasing in $x$ and the fact that $\displaystyle \lim_{x \to \infty} \Big(1+\frac 1x\Big)^{x+\alpha} = \lim_{x \to \infty} \Big(1+\frac 1x\Big)^x \lim_{x \to \infty} \Big(1+\frac 1x\Big)^\alpha =e$

provided $\alpha \geq 1/2$, we always have $\displaystyle e \leq \Big( 1 + \frac 1n \Big)^{n+\frac 12} \leq \cdots \leq \Big (1 + \frac 1n \Big)^{n+1}.$

Notice that generally the inequities $\displaystyle \Big( 1 + \frac 1n \Big)^n \leq \cdots \leq \Big( 1 + \frac 1n \Big)^{n+\frac 14} \leq \cdots \leq e$

do not hold if $n$ is small. For example $\displaystyle \Big( 1 + \frac 12 \Big)^{2+0.495} - e = 0.031813140>0.$

However, the above inequalities hold starting from large $n$.

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