Proving the inequality $\left(1+\frac{1}{n}\right)^x>1+\frac{x}{n+1}$ I want to prove that $\left(1+\frac{1}{n}\right)^x>1+\frac{x}{n+1}$ $\forall x \ge1$ and $n \in \mathbb{N}$.
I first expanded $\left(1+\frac{1}{n}\right)^x=1+\frac{x}{n}+\frac{x(x-1)}{2n^2}+...$. So I wanted to truncate the expansion but I think the terms start becoming negative later on in the series. So I do not know how to do it. If anyone can give any hints it would be great. Thanks.
 A: Hint: Prove instead the more attractive $(1+t)^x \ge 1+tx$ for $x\ge 1$ and $t\ge 0$.  This implies the desired result.
Let $f(t)=(1+t)^x - (1+tx)$. Then $f(0)=0$. Show that $f'(t)\ge 0$ for $t\ge 0$. 
A: This is direct using Generalised Bernoulli Inequality.  
If you are not familiar with that, here is a quick proof using Jensen's inequality and strict concavity of $\log$ function:
$$\frac1x \log\left(1+\frac{x}n\right)+\left(1-\frac1x\right)\log 1 < \log \left(1+\frac1n\right)$$
$$\iff \left(1+\frac1{n}\right)^x> 1+\frac{x}n >1+\frac{x}{n+1}$$
A: Let $f(x)=(1+\frac{1}{n})^{x}$, let $g(x)=1+\frac{x}{n}$. It is clear that when $x=1$ we have
$$
f(1)=g(1)=1+\frac{1}{n}
$$
And for all $x\ge 1$ we have
$$
f'(x)=f(x)\log(1+\frac{1}{n})=(1+\frac{1}{n})^{x}\log(1+\frac{1}{n}), g'(x)=\frac{1}{n}
$$
I claim that $f'(x)\ge f'(1)\ge g'(x)$ for all $x\ge 1, n\ge 1$. In other words we want to show that
$$
(1+y)\log(1+y)-y\ge 0,\forall 0\le y\le 1
$$
But this can be proved by differentiating again with respect to $y$, after which we have
$$
1+\log(1+y)-1=\log(1+y)\ge 0
$$
and it is clear that for $y=0$ we obtain an equality. So the original statement is proved, because $g(x)> \frac{x}{n+1}$ at all times. 
A: Here are some hints. Look at $1+\frac{x}{n}+\frac{x(x-1)}{2n^{2}}+\cdots$ and compare this infinite sum to $1+\frac{x}{n+1}$. Do we know which of these two sum is bigger? Also, use the generalized binomial theorem, which can be found about halfway down on this page: http://en.wikipedia.org/wiki/Binomial_theorem.
