How to prove $(1+1/x)^x$ is increasing when $x>0$? Let $F(x)=(1+\frac{1}{x})^x$.
How do we prove $F(x)$ is increasing when $x>0$?
 A: Let $h(x) = \mathrm e^{-1/(1+x)}(1+1/x)$ and note that for $x > 0$,
$$
h(x) = \mathrm e^{-1/(1+x)}\cdot\frac{x+1}{x} > \left(1-\frac{1}{1+x}\right)\frac{x+1}{x} = 1.
$$
Now, let $g(x) = \log F(x)$ and note that 
$$
g'(x) = \log(1+1/x) - \frac{1}{1+x} = \log h(x) > \log 1 = 0\>,
$$
and so we are done.
A: It's enough to show $\ln(1+1/x)^x = x\ln(1+1/x)$ is increasing. Letting $h=1/x$, we can write the last expression as
$$\tag 1 \frac{\ln(1+h)-\ln 1}{h}.$$
We want to show $(1)$ increases as $h$ decreases. Now $(1)$ is just the slope of the line through $(1,\ln 1)$ and $(1+h,\ln(1+h)).$ And any concave function has the property that such slopes increase as $h$ decreases. Since $\ln x$ is concave, we're done.
A: $(1+1/x)^x$ is increasing equivalent to $x\cdot\ln(1+1/x)$ is increasing by taking natural logarithm. Which is equivalent to derivative being positive, which is showing that $h(x) = \ln(1+1/x) - 1/(1+x) > 0$.
And now notice $h'(x)=-1/[x\cdot(x+1)^2]$ is negative, so $h(x)$ is decreasing and minimal value is at $+\infty$, but $h(+\infty)=0$. So, $h(x) \ge 0$.
A: We have to prove that 
$$F(x) = \left(1+\frac{1}{x}\right)^x$$
is a strictly increasing function for $x>0$.
The derivative is
$$F'(x) = \frac{\left(\frac{1}{x}+1\right)^x \left((x+1) \log \left(\frac{1}{x}+1\right)-1\right)}{x+1}$$
Hence we only need to show that
$$\log \left(\frac{1}{x}+1\right)-\frac{1}{x+1}>0 $$
But this is obvious since the l.h.s. is equal to the definitely positive integral
$$\int_x^{\infty } \frac{1}{t (t+1)^2} \, dt$$
A: There's an elementary approach for rational $x$. It suffices to prove that 
$$\left( 1+\frac{m}{n} \right)^n < \left( 1+\frac{m}{n+1} \right)^{n+1}$$ 
for $m,n$ positive integers. Whenever $0 \leq a < b$, we have $\frac{b^{n+1} - a^{n+1}}{b-a} = \sum_{k=0}^n a^{n-k}b^k < (n+1)b^n$ which rearranges to
$$[(n+1)a - nb] \cdot b^n < a^{n+1}.$$
Substituting $a = 1+m/(n+1)$ and $b = 1 +m/n$ into the above, the term in square braces (miraculously) reduces to $1$ and we get the desired bound. This is adapted from Foundations of Mathematical Analysis by Johnsonbaugh and Pfaffenberger.
A: $f(t):=\frac{1}{t}\log(1+t)$ is decreasing for $t>0$, because it is smooth and its derivative is negative.
Its derivative is $f'(t)=-\frac{1}{t^2}g(t)$, where $g(t):=\log(1+t)-\frac{t}{1+t}.$
But, $g(t)>0$ for $t>0$, in fact, $\lim_{t\to 0}\ g(t)=0$ and $g$ is increasing.
$g$ is increasing because it is smooth and its derivative is positive: $g'(t)=\frac{t}{(1+t)^2}$.
A: This proof is from theorem 140 from Hardy's Inequalities.
Let $f(x) = \ln\left[\left(1 + \frac{1}{x}\right)^x\right] = x(\ln(x+1) - \ln(x)).$ We refer to the mean value theorem: for each differentiable $g$,
$$
g(x + h) - g(x) = hg'(x + \theta h)
$$
for $\theta \in (0,1)$.
Applying the MVT to $g(x) = \ln(x)$, we get $\ln(x+1) - \ln(x) = \frac{1}{x + \theta}$. Since, $\frac{1}{x+\theta} > \frac{1}{x+1} $, we have that $\ln(x+1) - \ln(x) > \frac{1}{x+1}$, so
$$
f'(x) = \ln(x+1) - \ln(x) - \frac{1}{x+1} > 0.
$$
Hence, $f(x)$ and $e^{f(x)} = \left(1+\frac{1}{x}\right)^x$ are increasing with $x$.
A: The given inequality is equivalent to
$$ \ln (x+1)-\ln(x) > \frac{1}{x+1},\ \ \ \ \ \ \ \ \ \  x\gt 0.$$
Let $f(t)= \frac{1}{t}$. Then,
$$ \ln (x+1)-\ln(x) = \int_1^{x+1} f(t)\  dt - \int_1^x f(t)\  dt $$$$
\ \ \ \ \ =\int_x^{x+1} f(t)\  dt $$$$\ \ \ \ \ge 1\cdot f(x+1) $$$$\ \ \ \ \ \ \ \ \ \ \ \  \ =f(x+1) = \frac{1}{x+1}$$
since $f$ is a decreasing function.
The strict inequality can be obtained with a little more work; by breaking up the interval $[x, x+1]$ into two pieces -  such as $\left[x, x+{1\over 2}\right]$ and $\left[x+{1\over 2}, x+1\right]$  - and applying the same argument as above on each interval.
