How to show $|e^{it}-\sum_{k=0}^{k=n}\frac{(it)^k}{k!}|\leq \frac{|t|^{n+1}}{(n+1)!}$

Question

$$\left|e^{it}-\sum_{k=0}^{n}\frac{(it)^k}{k!}\right|=\left|\sum_{k=n+1}^{\infty}\frac{(it)^k}{k!}\right|=\frac{|t|^{n+1}}{(n+1)!}\left|\sum_{k=0}^{\infty}\frac{(it)^k}{(n+2)\cdots(n+1+k)}\right|$$

Is the term inside the modulus somehow less or equal to $1$?

Answer 1

You can use $$ \mathrm{e}^{\mathrm{i}t} - \sum\limits_{k = 0}^n {\frac{{(\mathrm{i}t)^k }}{{k!}}} = \frac{{(\mathrm{i}t)^{n + 1} }}{{n!}}\int_0^1 {\mathrm{e}^{\mathrm{i}tx} (1 - x)^n \mathrm{d}x} $$ for any real $t$. Then \begin{align*} \left| {{\rm e}^{{\rm i}t} - \sum\limits_{k = 0}^n {\frac{{({\rm i}t)^k }}{{k!}}} } \right| &= \frac{{\left| t \right|^{n + 1} }}{{n!}}\left| {\int_0^1 {{\rm e}^{{\rm i}tx} (1 - x)^n {\rm d}x} } \right| \\ & \le \frac{{\left| t \right|^{n + 1} }}{{n!}}\int_0^1 {\left| {{\rm e}^{{\rm i}tx} } \right|(1 - x)^n {\rm d}x} \\ & = \frac{{\left| t \right|^{n + 1} }}{{n!}}\int_0^1 {(1 - x)^n {\rm d}x} = \frac{{\left| t \right|^{n + 1} }}{{(n + 1)!}}. \end{align*}