How to show that $\left|\sum_{k=0}^\infty\frac{(ix)^k}{(k+1)!}\right|\le \left|\sum_{k=0}^\infty\frac{(ix)^k}{k!}\right|=|e^{ix}|=1$ with restrictions 
How to show that $\left|\sum_{k=0}^\infty\frac{(ix)^k}{(k+1)!}\right|\le \left|\sum_{k=0}^\infty\frac{(ix)^k}{k!}\right|=|e^{ix}|=1$ with restrictions, for $x\in\Bbb R$.

To prove this inequality we cant use any related to derivatives, integrals, geometric statements about sine or cosine, or uniform convergence. We can use limits and basic facts about the convergent properties of these power series.
We already knows that $|e^{ix}|=1$ for $x\in\Bbb R$. The inequality is a slight rewrite of
$$\frac{|e^{ix}-1|}{|x|}=\left|\sum_{k=0}^\infty\frac{(ix)^k}{(k+1)!}\right|\le 1,\quad\forall x\in\Bbb R$$
what need to be proved. I dont know exactly what to do here, Im completely lost. The best I can think is to prove something like
$$\forall\epsilon>0,\exists N\in\Bbb N:\left|\sum_{k=0}^n\frac{(ix)^k}{(k+1)!}-L\right|<\epsilon,\quad\forall n\ge N$$ for some $0\le L<1$. 
The exercise leave the hint $\lim_{z\to 0}\frac{\exp(z)-1}{z}=1$ for $z\in\Bbb C\setminus\{0\}$, but I dont see how to relate this to our problem, because we need the result for any $x$, not just for $x=0$. Some hint or solution will be appreciated, thank you.
 A: Note: This answer uses the identity $$e^{i(x+y)} = e^{ix} e^{iy}$$ which can be proved using Cauchy's product.

As the hint suggests, we have
$$\lim_{r \to 0} \frac{e^{ir}-1}{ir}=1. $$
For fixed $\epsilon>0$ we can choose $\delta>0$ such that
$$|e^{ir}-1| \leq (1+\epsilon) |r| \qquad \text{for all $|r| \leq \delta$}. \tag{1}$$
Now pick $x \in \mathbb{R}$ and choose $n \in \mathbb{N}$ sufficiently large such that $r := x/n$ satisfies $|r| \leq \delta$. For $$x_j := j r, \qquad j=0,\ldots,n$$ we have
$$e^{ix}-1 = \sum_{j=1}^n (e^{ix_j}-e^{ix_{j-1}}). \tag{2}$$
Using that
$$e^{ix_j}-e^{ix_{j-1}} = e^{ix_{j-1}} (e^{ir}-1)$$
and $|e^{ix_{j-1}}| \leq 1$, we get
$$|e^{ix}-1| \leq \sum_{j=1}^n |e^{ix_j}-e^{ix_{j-1}}| \leq n |e^{ir-1}| \stackrel{(1)}{\leq} (1+\epsilon) n \cdot |r| = (1+\epsilon) |x|.$$ Since $\epsilon>0$ was arbitrary, this gives
$$|e^{ix}-1| \leq |x| \qquad \text{for all $x \in \mathbb{R}$.}$$
Rewriting this identity using the definition of $e^{ix}$, this proves the assertion.
A: Just for the record I will add a second proof. We knows that
$$e^{ix}=\cos(x)+i\sin(x)$$
and $|e^{ix}|=1$ for all $x\in\Bbb R$. And we want to prove
$$\frac{|e^{ix}-1|}{|x|}\le 1$$
From the last inequality we have the bound
$$|e^{ix}-1|\le |e^{ix}|+1=2\le|x|$$
then for $|x|\ge 2$ the inequality is clear. Now, we will study the case for $|x|<2$. From the Euler's formula we have that
$$\frac{|e^{ix}-1|}{|x|}=\frac{|\cos(x)+i\sin(x)-1|}{|x|}=\\=\frac{\sqrt{(1-\cos (x))^2+\sin^2(x)}}{|x|}=\frac{\sqrt{2(1-\cos(x))}}{|x|}\le 1$$
Then our inequality can be written as
$$1-\cos(x)=\sum_{k=1}^\infty(-1)^{k+1}\frac{x^{2k}}{(2k)!}\le \frac{|x|^2}2\implies\sum_{k=2}^\infty(-1)^k\frac{x^{2k}}{(2k)!}\ge 0$$
Now observe that for $|x|<2$ and $k\ge 2$ the sequence $(x^{2k}/(2k)!)$ decreases monotonically. Then for these alternating series we have the bound
$$|s-s_n|\le |c_{n+1}|$$
where $s:=\sum_{k=0}^\infty (-1)^k c_k$ is an alternating series where $(c_n)\to 0$ monotonically, $s_n$ is a partial sum of the series, and $c_{n+1}$ is an element of the monotone sequence.
Then
$$\left|\sum_{k=2}^\infty(-1)^k\frac{x^{2k}}{(2k)!}-\frac{x^4}{4!}\right|\le\frac{x^6}{6!}\implies \sum_{k=2}^\infty(-1)^k\frac{x^{2k}}{(2k)!}\ge \frac{x^4}{4!}-\frac{x^6}{6!}=\frac{x^4}{4!}\left(1-\frac{x^2}{30}\right)\ge 0$$
whenever $|x|<2$.$\Box$
