Estimating the remainder term in $(1+x)^s$, $s\in\mathbb R^+$ If $s \in \mathbb R^+$ and $-1<x<1$, Then we have:
$$(1+x)^s=1+{s\choose 1}x+{s\choose 2}x^2+...+{s\choose n}x^n+R_{n+1}$$
Where $R_{n+1}$ is the remainder term, and $${s\choose k}=\prod_{i=0}^{k-1}\frac{s-i}{k}, k>0$$
I'm having a hard time trying to estimate the remainder term, it should be the usual integral (Taylor Series) but I don't know how to estimate it.
Edit:
since $\mid(1+x)\mid^s<2^s$, $\forall s\in \mathbb R^+$ so $$\mid R_n\mid\le 2^s \frac{\mid x\mid^n}{n!}$$
is that correct?
 A: We can write our formula succinctly in terms of the Gamma function. Given $|x|<1$ we can write
$$(1+x)^p=\sum_{k=0}^\infty \frac{\Gamma(p+1)}{\Gamma(p-k+1)}\frac{x^k}{k!}$$
Let's stop the sum at some positive integer $n$, in particular, let $n=\lfloor p\rfloor$.
$$(1+x)^p=\sum_{k=0}^n \frac{\Gamma(p+1)}{\Gamma(p-k+1)}\frac{x^k}{k!}+\sum_{k=n+1}^\infty \frac{\Gamma(p+1)}{\Gamma(p-k+1)}\frac{x^k}{k!}$$
The $\Gamma(p+1)$ is a constant. Let's write this as
$$(1+x)^p=\sum_{k=0}^n \frac{\Gamma(p+1)}{\Gamma(p-k+1)}\frac{x^k}{k!}+\Gamma(p+1)\sum_{k=n+1}^\infty a_kx^k$$
With $a_k=\left(\Gamma(p-k+1)k!\right)^{-1}$. Our task now is to show that $|a_k|$ is decreasing.
We have
$$a_{k+1}=\frac{1}{\Gamma(p-k-1+1)(k+1)!}$$
$$=\frac{1}{\Gamma(p-k)k!}\cdot\frac{1}{k+1}=\frac{1}{\Gamma(p-k+1)k!}\cdot\frac{p-k}{k+1}=\frac{p-k}{k+1}a_k$$
Since $k\geq n+1=\lfloor p\rfloor+1$, $|(p-k)/(k+1)|\leq 1$. Hence $|a_k|$ is decreasing. So, we can bound the sum by the first coefficient:
$$\left|\sum_{k=n+1}^\infty a_kx^k\right|\leq\left|\sum_{k=n+1}^\infty a_{n+1}x^k\right|$$
But this sum is now easy, and can be evaluated with a geometric series (again remember $|x|<1$). So
$$\left|\sum_{k=n+1}^\infty a_{n+1}x^k\right|=\left|a_{n+1}x^{n+1}\sum_{k=0}^\infty x^k\right|=\left|\frac{x^{n+1}}{\Gamma(p-k)(n+1)!(1-x)}\right|$$
So going back to our original expansion
$$(1+x)^p=\sum_{k=0}^n \frac{\Gamma(p+1)}{\Gamma(p-k+1)}\frac{x^k}{k!}+\underbrace{\Gamma(p+1)\sum_{k=n+1}^\infty a_kx^k}_{R_{n+1}}$$
We have
$$|R_{n+1}|\leq\left|\frac{\Gamma(p+1)x^{n+1}}{\Gamma(p-n)(n+1)!(1-x)}\right|$$
You can check this numerically on Desmos.
