How do I prove this Taylor Series Problem? Let $p\in \mathbb{R}$, and let $f(x) = (1 + x)^p$ for $-1 < x < +\infty$. Show that the Taylor series $\sum_{k=0}^{\infty} \frac{f^{(k)}(0) x^{k}}{k!}$ converges to $f(x)$ in the interval $-1 < x < 1$.
I have learned a littler bit of Taylor Series in my engineering courses but have not learned these proof type questions before. Can someone please help? This is my first attempt at learning analysis. 
 A: We need to use Cauchy's form of remainder in Taylor's theorem :

Taylor's Theorem with Cauchy's Remainder: If the $n$-th derivative $f^{(n)}(x) $ exists in an open interval $I$ containing point $a$ then for all values of $h$ such that $a+h\in I$ we have a corresponding number $\theta\in(0,1)$ such that $$f(a+h) =f(a) +hf'(a) +\frac{h^{2}}{2!}f''(a)+\dots+\frac{h^{n-1}}{(n-1)!}f^{(n-1)}(a)+R_{n}$$ where $$R_{n} =\frac{(1-\theta)^{n-1}f^{(n)}(a+\theta h) h^{n} } {(n-1)!} $$

For the current question let $f(x) = (1+x)^{p},a=0,h=x$ and note that $$f^{(n)} (x) =p(p-1)\dots(p-n+1)(1+x)^{p-n}$$ Then we have via above mentioned theorem $$(1+x)^{p}=\sum_{k=0}^{n-1}\binom{p}{k}x^{k}+R_{n}\tag{1}$$ where $$R_{n} =\frac{p(p-1)\dots(p-n+1)} {(n-1)!}\cdot\frac{(1-\theta)^{n-1}x^{n}}{(1+\theta  x) ^{n-p}} \tag{2}$$ If $p$ is a positive integer then $R_{p+1}=0$ so that the series is finite with sum $(1+x)^{p}$ (this is the usual binomial theorem for positive integer index $p$). If $p$ is not a positive integer then we need to analyze the expression $R_{n} $ a bit more carefully. 
If $|x|<1$ then $0<(1-\theta)/(1+\theta x) <1$ and $(1+\theta x)^{p-1}$ is less than $(1+|x|)^{p-1}$ if $m>1$ and less than $(1-|x|)^{p-1}$ if $p<1$. Therefore $$|R_{n} |<|p|(1\pm|x|)^{p-1}\left|\binom {p-1}{n-1}\right||x|^{n}\tag{3}$$ where the sign $+$ or $-$ is chosen according as $p>1$ or $p<1$.
It can be easily proved (via ratio test) that the RHS of the above inequality tends to $0$ as $n\to\infty$ provided that $|x|<1$. Thus $R_{n} \to 0$ as $n\to\infty$ and therefore from equation $(1)$ it follows that the series in question converges to sum $(1+x)^{p}$ for $|x|<1$. Things are a bit tricky when $x=\pm 1$. More details are available in my blog posts here and here.
A: Recall  Taylor-Young formula asserts that
$$ (1+x)^p=\sum_{k=0}^{n} \frac{f^{(k)}(0) x^{k}}{k!}=\sum_{k=0}^{n} \frac{p(p-1)\dotsm(p-k+1)}{k!}x^{k}+R_n(x) $$
where  $\quad R_n(x)= o(x^n)$. There results $R_n(x)\to 0$ as $n\to\infty$ if $\,|\,x\,|<1$.
