Uniform convergence of $\sin x$ - what to choose for $N$? I'm trying to show uniform convergence of the power series of $\sin x$ on $[0,r] (r>0)$ from first principles. So let $\epsilon>0$ and $n\geq N$. I'm stuck at this step: What do I choose for $N$ (a function of $\epsilon$) so that 

$$\sum\limits_{k=n+1}^\infty\frac{r^{2k+1}}{(2k+1)!} <\epsilon ?$$

(I know the usual method is to observe that the LHS expression above converges to $0$, but here I would like to use a pure $N-\epsilon$ argument.)
 A: Using part of Stirling's approximation implying that $k!>\left(\frac{k}{e}\right)^k$ allows us to get a crude geometric series bound when $n$ is big enough:
$$
\begin{align*}
\sum\limits_{k=n+1}^\infty\frac{r^{2k+1}}{(2k+1)!} 
&=r\sum\limits_{k=n+1}^\infty\frac{(r^2)^k}{(2k+1)!}\\
&< r\sum\limits_{k=n+1}^\infty\frac{(r^2)^k}{k!}\\
&<r\sum\limits_{k=n+1}^\infty\left(\dfrac{er^2}{k}\right)^k\\
&<r\sum\limits_{k=n+1}^\infty\left(\dfrac{er^2}{n+1}\right)^k.\\
\end{align*}
$$
The geometric series in the last expression could diverge to $+\infty$.  However, it is always okay to add assumptions that require $n$ to be larger.  In particular, if $n+1>er^2$, then the geometric series converges, showing that 
$$\sum\limits_{k=n+1}^\infty\frac{r^{2k+1}}{(2k+1)!}<\dfrac{r\left(\frac{er^2}{n+1}\right)^{n+1}}{1-\frac{er^2}{n+1}}.$$
In order to get a bound that is easier to work with, we can use the inequality $\dfrac{a^k}{1-a}<2^{k-1}$ if $0<a<\frac12$.  So making the stronger assumption that $n+1>2er^2$ implies that 
$$\sum\limits_{k=n+1}^\infty\frac{r^{2k+1}}{(2k+1)!}<r2^{-n}.$$
We easily see that $n>\dfrac{\ln(r/\varepsilon)}{\ln(2)}$ implies that $r2^{-n}<\varepsilon$, so under the standing assumption that $n+1>2er^2$, the inequality $n>\dfrac{\ln(r/\varepsilon)}{\ln(2)}$ implies that $\sum\limits_{k=n+1}^\infty\frac{r^{2k+1}}{(2k+1)!} <\epsilon$.  
Thus $N>\max\left\{\dfrac{\ln(r/\varepsilon)}{\ln(2)},2er^2\right\}$ suffices, because any $n>N$ automatically satisfies $n+1>2er^2$, so that the bound $r2^{-n}$ for our series applies.
