Show that $n^{\frac{n+1}{n}}$ is increasing How can one show that $n^{\frac{n+1}{n}}$ is increasing, possibly using logarithms but not derivatives?  I obtain $(n^2+2n)log(n+1)\geq (n^2+2n+1)log (n)$ but cannot proceed further. 
 A: Hint : Use the fact that $\ln(n+1) \geq \ln(n) +\frac{1}{n+1}$
Lemma : $\ln(n+1) \geq \ln(n)+\frac{1}{n+1}$
Or we have $\ln(\frac{n+1}{n}) \geq \frac{1}{n+1}$
Exponent-ate both side gives us that $1+\frac{1}{n} =1+\frac{1}{n+1}+\frac{1}{(n+1)^2}+\cdots\geq e^{\frac{1}{n+1}} = 1+\frac{1}{n+1}+\frac{1}{2(n+1)^2}+\cdots$ 
Visually the proof is obvious.
A: A purely algebraic proof
We have to prove that
$$\bigl(n+1\bigr)^{\tfrac{n+2}{n+1}}> n^{\tfrac{n+1}{n}}\iff (n+1)^{n(n+2)}>n^{(n+1)^2}.$$
Note that $\;(n+1)^{n(n+2)}=(n+1)^{(n+1)^2-1} $, so the last inequality is equivalent to
$$\frac{(n+1)^{(n+1)^2-1}}{n^{(n+1)^2}}>1\iff\Bigl(1+\frac1n\Bigr)^{(n+1)^2}>n+1$$
Let's apply Bernoulli's inequality:
$$\Bigl(1+\frac1n\Bigr)^{(n+1)^2}>1+\frac{(n+1)^2}{n}=1+n+2+\frac 1n>3+n.$$
A: How about this: 
Since $n^{1+1/n}=e^{(1+1/n)\ln{n}}$, and the exponential is a monotonically increasing function, so is $(1+1/n)\ln{n}$. Then verify that the function $g(x)=(1+1/x)\ln x$ is monotonically increasing.
One has only to prove that $g'(x)>0$ $\forall x>0$.
We have
\[
g'(x)= -\frac{1}{x^2}\ln x+\left(1+\frac{1}{x}\right)\frac{1}{x}
\]
Then, using the inequality $-\ln x\geq 1-x$, valid for $x>0$, we obtain
\[
g'(x)\geq (1-x)\frac{1}{x^2}+\left(1+\frac{1}{x}\right)\frac{1}{x}=\frac{2}{x^2}>0
\]
A: Note (added after posting):  I didn't notice the OP's proscription on derivatives. I'm leaving the answer in case anyone wants to see an approach ignoring the proscription.
Let $u=1/n$.  Then
$$n^{n+1\over n}=n^{1+{1\over n}}=\left(1\over u\right)^{1+u}={1\over u^{1+u}}$$
It suffices to show that $u^{1+u}$ is an increasing function of $u$, which is equivalent to showing that $f(u)=(1+u)\ln u$ is an increasing function:
$$f'(u)={1+u\over u}+\ln u={1+u\over u}+\int_1^u{dt\over t}\ge{1+u\over u}+\int_1^u{dt\over u}={1+u\over u}+{u-1\over u}=2\gt0$$
Remark:  The inequality 
$$\int_1^u{dt\over t}\ge\int_1^u{dt\over u}$$
is easy to see for $u\ge1$, since $1/t$ is being replaced by its smallest value, $1/u$.  For $0\lt u\lt1$ it's a bit trickier:  You're replacing $1/t$ with its largest value, but you're integrating "backwards," producing a negative number.
