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

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