Show that $\lvert\frac{1}{x} \sin x-\sum _{k=0}^n (-1)^k\frac{x^{2k}}{(2k+1)!}\rvert\leq\frac{|x|^{2n+1}}{(2n+2)!}$

Question

Prove that

$$\left\lvert\frac{1}{x} \sin x-\sum _{k=0}^n(-1)^k \frac{x^{2k}}{(2k+1)!}\right\rvert\leq\frac{|x|^{2n+1}}{(2n+2)!}$$

I expanded $\dfrac{1}{x} \sin x$ and I got it to be $\displaystyle\sum_{k=0}^\infty (-1)^k\frac{x^{2k}}{(2k+1)!}$.

But how can I show this inequality?

Should I use induction ?

Answer 1

Hint. By using the Lagrange remainder form of the Taylor expansion of $f(x)=\sin(x)$ at $x_0=0$ we have that for any $x\not=0$, there is $t$ with $0<|t|<|x|$ such that $$\sin x=\sum _{k=0}^n \frac{(-1)^kx^{2k+1}}{(2k+1)!}+\frac{f^{2n+2}(t)x^{2n+2}}{(2n+2)!}.$$