Binomial Expansion, Taylor Series, and Power Series Connection 1) Is there a reason why the binomial expansion of $(a+x)^n$ is the same as a Taylor series approximation of $(a+x)^n$ centered at zero? 
2) The binomial expansion of $(a+x)^n$ is 
$a^n + na^{n-1}x + \frac{n(n-1)}{2!}a^{n-2}x^2 +$.... 
If the expansion is written this way, then $n$ can be an integer (positive or negative) or a fraction? 
If the binomial expansion is written in summation notation using nCr, then n can only be positive because nCr cannot have a negative $n$? 
3) For the expansion of $(a+x)^n$ I gave in question 2, does $a$ have to be $a = 1$ with $-1 < x < 1$? 
What are these restrictions? 
Update : An infinite geometric series converges when the common ratio, $x$ in this case, is between -1 and 1. The infinite binomial expansion I wrote in question 2 is a valid expansion of $(a+x)^n$ when $-1 < x < 1$. So if I put $x = 0.5$ into $(a+x)^n$ for a given $n$ and $a$, $(a+0.5)^n$ and the infinite expansion for $x = 0.5$ will give the same answer. If I use $x = 40$, the expansion will diverge and not give the same answer as the original function $(a+40)^n$. Does this mean that the binomial expansion is actually a power series (a geometric series is a special case of a power series)? And does $a$ need to be 1 for the $-1 < x < 1$ to be required? Or is it required regardless of what $a$ is?
 A: 1) They are the same function, so they have the same power series.
2)
In this answer, it is shown that for the generalized binomial theorem, we have for negative exponents,
$$
\binom{-n}{k}=(-1)^k\binom{n+k-1}{k}
$$
Thus, we have
$$
\begin{align}
(a+x)^{-3}
&=a^{-3}\left(1+\frac xa\right)^{-3}\\
&=a^{-3}\sum_{k=0}^\infty\binom{-3}{k}\left(\frac xa\right)^k\\
&=a^{-3}\sum_{k=0}^\infty\binom{k+2}{k}\left(\frac xa\right)^k\\
&=\sum_{k=0}^\infty\binom{k+2}{2}\frac{x^k}{a^{k+3}}\\
\end{align}
$$
The same can be done for fractional exponents, but the formulas for the coefficients are more complicated.
3) In the answer to 2), we factored out the $a^{-3}$ so that one term of the sum was $1$. This allows us to use the binomial theorem in an open-ended way; that is, we don't need to worry about what the exponent of $n-k$ needs to be. In particular, the generalized binomial theorem reads
$$
(1+x)^n=\sum_{k=0}^\infty\binom{n}{k}x^k
$$
where
$$
\binom{n}{k}=\frac{n(n-1)(n-2)\dots(n-k+1)}{k!}
$$
Furthermore, if $n$ is not a non-negative integer, the binomial expansion does not terminate. In that case, the series for
$$
(a+x)^n=a^n\left(1+\frac xa\right)^n
$$
converges for $|x|\lt|a|$.

Extension for $\boldsymbol{|x|\gt|a|}$
We can extend the convergence of a series for $(a+x)^n$ for $|x|\gt|a|$ if we allow Laurent expansions and write
$$
(a+x)^n=x^n\left(1+\frac ax\right)^n
$$
Using the same example as above,
$$
\begin{align}
(a+x)^{-3}
&=x^{-3}\left(1+\frac ax\right)^{-3}\\
&=x^{-3}\sum_{k=0}^\infty\binom{-3}{k}\left(\frac ax\right)^k\\
&=x^{-3}\sum_{k=0}^\infty\binom{k+2}{k}\left(\frac ax\right)^k\\
&=\sum_{k=0}^\infty\binom{k+2}{2}\frac{a^k}{x^{k+3}}\\
\end{align}
$$
which converges for $|x|\gt|a|$.
