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?

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
$$
(1-|x|)^n\ge 1-n|x|>1-2n|x|=1+2nx.
$$
