An inequality with $e^{ix}$

Question

I am to prove the following statement $$\bigg|e^{ix} - \sum_{k=0}^m \frac{(ix)^k}{k!} \bigg| \le \frac{|x|^{m+1}}{(m+1)!},$$ where $x \in \mathbb{R}$.
I used Taylor's expansion of $e^x$. That led me to this $$\bigg|\sum_{k=m+1}^{\infty}\frac{(ix)^k}{k!} \bigg| \le \frac{|x|^{m+1}}{(m+1)!}$$ Let's focus on the LHS $$\bigg|\sum_{k=m+1}^{\infty}\frac{(ix)^k}{k!} \bigg|\le \sum_{k=m+1}^{\infty}\big|\frac{(ix)^k}{k!} \big| = \sum_{k=m+1}^{\infty}\frac{|ix|^k}{k!}$$ Because $|i| = 1$ we get $$\sum_{k=m+1}^{\infty}\frac{|x|^k}{k!}$$ Let's look at the RHS. It's the first component of the sum on LHS. All the elements of the sum are positive thus the first one cannot be bigger then the sum. What have I done wrong?

Answer 1

Hint: Try Taylor's theorem with the integral form of the remainder, which works for complex valued functions: If $f:\mathbb R\to \mathbb C$ is infinitely differentiable, then for $m=1,2,\dots,$

$$f(x) = \sum_{k=0}^{m}\frac{D^kf(0)}{k!}x^k +R_m(x),$$where $$ R_m(x)= \int_0^x\int_0^{t_1}\cdots \int_0^{t_m}D^{m+1}f(t_{m+1})\,dt_{m+1}\, dt_{m}\,\dots dt_{1}.$$