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.
 A: 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,
$$
\frac{{\rm d}}{{{\rm d}u}}\frac{u}{{1 + u}} = \frac{1}{{(1 + u)^2 }} \le \frac{1}{{1 + u}} = \frac{{\rm d}}{{{\rm d}u}}\ln (1 + u).
$$
A: 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
$$ (1 + ax)^{\frac{1}{a}} \ge (1 + bx)^{\frac{1}{b}}$$
