I have to show that sequences $\left(\frac{x^{2k+1}}{(2k+1)!}\right)$ where $x\in(0,1)$ and $\left(-\frac{x^{2k+1}}{(2k+1)!}\right)$ where $x\in(-1,0)$ decrease monotonically and converge to $0$.
I know that if $x\in(0,1)$ then the sequnce $(x^n)$ converges to $0$ but in this case I have the $(2k+1)!$
It will enough to show that $$\frac{x^2}{(2k+2)(2k+3)}<1\qquad\text{for }0<x<1$$ Which is clear. So $$\frac{x^{2(k+1)+1}}{(2(k+1)+1)!}=\frac{x^2}{(2k+2)(2k+3)}\cdot\frac{x^{2k+1}}{(2k+1)!}<\frac{x^{2k+1}}{(2k+1)!}$$ Then, the sequence is convergent.
On the other hand, the following argument shows that the sequence converges to zero:
Let $k\ge 1$ an integer, then for $-1<x<1$ we have $$0\leq\left|\frac{x^{2k+1}}{(2k+1)!}\right|\leq|x^{2k+1}|$$ And $|x^{2k+1}|_{k\in\mathbb{N}}$ converges to $0$.
Hint: $$\frac{|x|^{2k+1}/(2k+1)!}{|x|^{2k}/(2k)!}=\frac{|x|}{2k+1}<1\ \forall k\ge 0, \mathrm{if}\ x\in (-1,1)$$
