Help in proving an inequality with logarithms I want to show that:
$(n+1)(\log(n+1)-\log(n)) > 1$, if $n>0$.
I have tried exponentiating it and I got
$( (n+1)/n )^{n+1} > e.$
Then I tried to show that the limit of  $( (n+1)/n )^{n+1}$ as $n\to\infty$ is $e$, and knowing that at $n=1$, it is greater than $e$. 
So, what's left is to show that $( (n+1)/n )^{n+1}$ is monotonically decreasing, but I to evaluate the sign of its derivative, I've got to prove that:
$n \log((n+1)/n) < 1$ for all $n>0$.
So, right now it seems like I'm going in circles. Could anyone help me please?
 A: As for $x >0$ $f(x)=\frac{1}{x}$ is non increasing, you have for $n < x < n+1$ $$\frac{1}{n+1} < \frac{1}{x}$$ You can now integrate this inequality between $n$ and $n+1$ to get the desired result as $f$ is continuous.
A: Consider the function
$$
f(x)=(x+1)(\log(x+1)-\log x)
$$
defined for $x>0$. The derivative is
$$
f'(x)=\log(x+1)-\log x+(x+1)\left(\frac{1}{x+1}-\frac{1}{x}\right)
=\log(x+1)-\log x-\frac{1}{x}
$$
that can be rewritten as
$$
f'(x)=\log\left(1+\frac{1}{x}\right)-\frac{1}{x}
$$
Now, for $t>0$, we have $\log(1+t)<t$, which can be easily proved by showing the function $g(t)=\log(1+t)-t$ is decreasing:
$$
g'(t)=\frac{1}{1+t}-1=-\frac{t}{1+t}<0
$$
and $g(0)=0$.
Therefore $f'(x)<0$ and so $f$ is strictly decreasing. Since
$$
\lim_{x\to\infty}f(x)=1
$$
we are done.
A: Let's use algebraic-geometric inequality to show the sequence is monotonically decreasing, which is equivalent to show its reciprocal $\left(\frac{n}{n + 1}\right)^{n + 1}$ is increasing. Write
$$\left(\frac{n}{n + 1}\right)^{n + 1} = 1 \times \left(\frac{n}{n + 1} \right) \times \cdots \times\left(\frac{n}{n + 1} \right),$$
where there are $n + 1$ fraction terms on the right hand side. Apply AM-GM inequality 
$$a_1 \cdots a_{n + 2} \leq \left(\frac{a_1 + \cdots + a_{n + 2}}{n + 2}\right)^{n + 2}$$
to the right hand side, we obtain that (notice that the equality cannot be achieved so there holds strict inequality):
$$\left(\frac{n}{n + 1}\right)^{n + 1} < \left[\frac{1 + (n + 1) \frac{n}{n + 1}}{n + 2}\right]^{n + 2} = \left(\frac{n + 1}{(n + 1) + 1}\right)^{n + 2}.$$
The prove is complete.
A: $$\log\left(\frac{n+1}{n}\right)=\int_0^\frac{1}{n+1}\frac{1}{1-x}\,dx\ge \int_0^\frac{1}{n+1}1\,dx=\frac{1}{n+1}$$
$$\log(n+1)-\log(n)\ge \frac{1}{n+1}\implies (n+1)(\log(n+1)-\log(n))\ge1$$
