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).

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