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<x< a/(1-2a)\\ < 0 & \quad  \text{ if } a<1/2, \quad x \geq a/(1-2a).\end{aligned}\right.

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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