# Establishing the inequality at the heart of a popular sequence which converges to powers of $e$

I know that the sequence

$$\displaystyle (1+kx)^\frac{1}{k},$$

where the sequence $\{k_i\}$ converges to zero, converges to $e^x$.

I also know the sequence is increasing. How does one show this is increasing? I am interested in neat ways of establishing the inequality,

$(1+ax)^\frac{1}{a} \ge (1+bx)^\frac{1}{b}$ if $b \ge a$

rather than the sequence itself.

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This can be proved using the Bernoulli's inequality:

$$(1+y)^r \ge 1 + ry \ \text{ if } r \ge 1 \ \text{ and } y \gt -1$$

Let $b = au$ where $u \ge 1$, then we have that

$$(1+ ax)^{\frac{b}{a}} = ( 1 + ax)^{u} \ge 1 + aux = 1 + bx$$

thus

$$(1 + ax)^{\frac{1}{a}} \ge (1 + bx)^{\frac{1}{b}}$$

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Assume that $x > 0$. We want to show that the function $f$ defined by $$f(t) = (1 + tx)^{1/t} , \;\; t > 0,$$ is decreasing. It suffices to show that $\ln f(t)$ is decreasing. Now $$\ln f(t) = \frac{{\ln (1 + tx)}}{t},$$ and $$\frac{{\rm d}}{{{\rm d}t}} \frac{{\ln (1 + tx)}}{t} = \frac{{xt/(1 + tx) - \ln (1 + tx)}}{{t^2 }}.$$ So we want $$\frac{{xt}}{{1 + tx}} \le \ln (1 + tx),$$ or $$\frac{{u}}{{1 + u}} \le \ln (1 + u), \;\; u > 0.$$ Since both sides of this inequality are $0$ at $u=0$, it suffices to show that $$\frac{{\rm d}}{{{\rm d}u}}\frac{u}{{1 + u}} \le \frac{{\rm d}}{{{\rm d}u}}\ln (1 + u).$$ Indeed, $$\frac{{\rm d}}{{{\rm d}u}}\frac{u}{{1 + u}} = \frac{1}{{(1 + u)^2 }} \le \frac{1}{{1 + u}} = \frac{{\rm d}}{{{\rm d}u}}\ln (1 + u).$$

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Thanks for this answer. I learned some interesting tricks. –  Henry B. Jul 17 '11 at 18:28
Glad you find my answer useful. –  Shai Covo Jul 17 '11 at 18:30