Power series expansion for sine function

Question

I wonder if the power series expansion of sine is a Cauchy sequence, i.e. the sequence $(\sum_{i=0}^{j}\frac{(-1)^{i}}{(2i+1)!}r^{2i+1})_{i\in\mathbb{N}}$ is Cauchy? Let $\varepsilon>0$. Is it possible to find the very integer $N>0$ such that for any $m,n\geq N$ $$ \left|\sum_{j=n}^{m}\frac{(-1)^{j}}{(2j+1)!}r^{2j+1}\right|\leq\sum_{j=n}^{m}\frac{1}{(2j+1)!}r^{2j+1}<\varepsilon $$ where $0\leq r\leq2\pi$?

Answer 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.

Answer 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.

If you want me to spell it out to you more explicitly don't hesitate asking.