How to prove $(1 + x)^{n} \leq 1 + 2nx$ I am currently working on a math problem, and it boils down to proving $(1 + x)^{n} \leq 1 + 2nx$, for a small $x$
By the Binomial expansion, it is clear that $$(1 + x)^n= 1+ nx + \frac{n(n-1)}{ 2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots$$
However, how can we prove that $$nx \geq \frac{n(n-1)}{ 2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots $$
which would prove my claim.
Any ideas?
 A: Since $(1 + x)^n$ is convex for $x \ge -1$, it lies below all of its secant lines. This gives that for any $x_0 > 0$ we have
$$(1 + x)^n \le 1 + \frac{(1 + x_0)^n - 1}{x_0} x, x \in [0, x_0].$$
So it suffices to choose $x_0 > 0$ small enough such that $(1 + x_0)^n - 1 \le 2n x_0$. We can take $x_0 = \frac{1}{n}$, since $\left( 1 + \frac{1}{n} \right)^n - 1 < e - 1 < 2 = \frac{2n}{n}$, so we get the slightly stronger result
$$(1 + x)^n \le 1 + (e-1)nx, x \in \left[ 0, \frac{1}{n} \right].$$
A: $(1 + x)^n = 1 + nx + O(x^2) = 1 + 2nx(1/2 + O(x)) \leq 1 + nx$ when the $O(x)$ term is $\leq 1/2$. The reason for the first equality is that a polynomial $p(x)$ of order $n$ is $O(x^n)$ as $x \to 0$. This is easy to prove, because $p(x)/x^n \to a_n$ as $x \to 0$, so in fact $p(x) \sim a_nx^n$.
A: One can continue from the binomial expansion, for $x>0$ this gives estimates against a geometric series
\begin{align}
(1+x)^n&= 1+nx\left(1+\frac{(n-1)x}{2}+\frac{(n-1)(n-2)x^2}{2·3}+...\right)
\\
&\le 1+nx\left(1+\frac{(n-1)x}{2}+\frac{(n-1)^2x^2}{2^2}+...\right)
\\
&\le 1+\frac{nx}{1-\frac{(n-1)x}{2}}
\end{align}
For the convergence of the series one needs $\frac{(n-1)x}{2}<1$, then for the claim $\frac{(n-1)x}{2}\le\frac12$, thus for $n\ge 2$ and $0\le x\le\frac1{n-1}$ the claimed inequality is true, for $n=0,1$ it is true for all $x>0$.
For $x<0$ the claim is wrong, as by the Bernoulli inequality
$$
(1-|x|)^n\ge 1-n|x|>1-2n|x|=1+2nx.
$$
A: For $|x|<\frac \epsilon n$, $$\left|\frac{(n)_k}{k!}x^k\right|<\frac{\epsilon^{k-1}}{k!}|nx|,$$
where $(n)_k=n(n-1)\cdots(n-k+1)$. And thus the absolute value of RHS is smaller than $$\frac{|nx|}{\epsilon}\sum_{k\ge 2}\frac{\epsilon^k}{k!}=\frac{|nx|}{\epsilon}(e^{\epsilon}-1-\epsilon)\le \epsilon|nx|.$$
when $\epsilon$ is small enough.
