Consider the power series expansion of sine function $$S=(\sum_{i=0}^{j}\frac{(-1)^{i}}{(2i+1)!}r^{2i+1})_{i\in\mathbb{N}}, 0\leq r\leq2\pi.$$ Then is the series $S$ a Cauchy sequence?

Let $\varepsilon>0$. Is it possible to find the very integer $N>0$ such that for any $m,n\geq N$

$$\left|\sum_{j=n}^{m}\frac{(-1)^{j}}{(2j+1)!}r^{2j+1}\right|\leq\sum_{j=n}^{m}\frac{1}{(2j+1)!}r^{2j+1}<\varepsilon ?$$

Proof 1. It is well-known that the power series expansion of $\sin x$ is convergent everywhere (you can use the ratio test, for example, to prove that), and all convergent sequences are Cauchy, so yes.

Proof 2. Yes that is indeed the case, as you are looking at the series within a compact set. The easiest way to see that is to look at the following inequality:

$$\sum_{j=n}^{m}\frac{1}{(2j+1)!}r^{2j+1}\leq\sum_{j=n}^{m}\frac{1}{(2j+1)!}(2\pi)^{2j+1}<\varepsilon$$

for $n$ big enough. That is the case because the last series is simply a converging series (quotient test,...), that means its tail must get arbitrarily small.

Proof 3. Consider the following inequality:

$$\sum_{j=1}^{\infty}\frac{1}{(2j+1)!}r^{2j+1} \leq e^{r}<\infty$$

Can you give a complete Proof 3 ?🤔