Solution (a) The value of $f$ and its derivatives at $0$ are $$\begin{array}{rl@{\qquad}rll} f(x) &=\sin x & f(0) &=0 \\ f^{\prime} (x) &=\cos x & f^\prime (0) &=1 \\ f^{\prime \prime} (x)&=-\sin x & f^{\prime \prime} (0) &=0 \\ f^{\prime \prime \prime}(x) &=-\cos x & f^{\prime \prime \prime} (0) &=-1 \end{array}$$

Higher-order derivatives follow this same pattern, so if $f(x) =\sin x$ can be represented by a power series in $x,$ then $$\begin{eqnarray*} \sin x &=& f(0) +\frac{f^{\prime} (0) }{1!}x+\frac{ f^{\prime \prime} (0) }{2!}x^{2}+\frac{f^{\prime \prime \prime} (0) }{3!}x^{3}+\cdots\,=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}- \frac{x^{7}}{7!}+\cdots\\[4pt] &=& \sum\limits_{k\,=\,0}^{\infty }\,(-1) ^{k}\frac{x^{2k+1}}{(2k+1) !} \end{eqnarray*}$$

(b) To prove that the series actually converges to $\sin x$ for all $x$, we need to show that the remainders $R_{2n+1}(x)$ and $R_{2n}(x)$ approach zero. For $R_{2n+1}(x)$ we have $$R_{2n+1}(x)=(-1)^{n+1}\frac{f^{(2n+2)}(u)\,x^{2n+2}}{(2n+2)!}$$

where $u$ is between $0$ and $x$. Since $\vert f^{(2n+2)}(u)\vert = \vert \sin u \vert \leq 1$ for every number $u$, then $$0\leq \vert R\,_{2n+1}(x)\vert =\frac{\left\vert f^{(2n+2)}(u)\right\vert }{(2n+2)!}\vert x\vert ^{2n+2}\leq \frac{ \vert x\vert ^{2n+2}}{(2n+2)!}$$

By the Ratio Test, the series $\sum\limits_{k\,=\,0}^{\infty }\frac{|x|^{2k+2} }{( 2k+2) !}$ converges for all $x$, so $$\lim_{n\,\dot{\rightarrow}\infty }\frac{|x|^{2n+2}}{(2n+2)!}=0$$

By the Squeeze Theorem, $\lim\limits_{n\rightarrow \infty }$$\vert R_{2n+1}(x)\vert =0$ and therefore $\lim\limits_{n\rightarrow \infty }$ $R_{2n+1}(x)=0$ for all $x.$

A similar argument holds for the remainder $\left\vert R_{2n}(x) \right\vert =\frac{\left\vert \cos u\right\vert x^{2n+1}}{(2n+1) !}.$

We conclude that the series $\sum\limits_{k\,=\,0}^{\infty }\,(-1) ^{k}\frac{x^{2k+1}}{(2k+1) !}$ converges to $\sin x$ for all $x.$ That is, $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \sin x=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!} -\cdots +(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}+\cdots =\sum\limits_{k\,=\,0}^{\infty }\,(-1) ^{k}\frac{x^{2k+1}}{(2k+1)!} }}$$