What is the Taylor series of a square root? I recently learned more about Taylor series, what I called infinite polynomials, and decided to find the Taylor series of $\sqrt{x}$. Of course, because $\frac{d}{dx}\sqrt{x}$ at $x=0$ is undefined, I am actually asking about the Taylor series of $\sqrt{x+1}$. I have found the Taylor series for this, kinda. It is $\sum_{n=0}^∞\left(\left(\prod_{m=1}^n\left(1.5-m\right)\right)\cdot\frac{x^n}{n!}\right)$. My real question is: Can this be made smaller?
 A: Here is the "smallest" representation I know of.
$$\sqrt{x+1} = \sum_{n\ge 0} \dbinom{\tfrac{1}{2}}{n} x^n$$
This is not strictly a Taylor series. It is actually the Binomial Expansion. It is still an infinite series. Is this what you are looking for?
Edit: Actually, looking at the series you calculated, these two representations should be the same.
A: There  is a general formula for the expansion of a binomial: for any $\alpha\in\mathbf R$, one has
$$(1+x)^\alpha=1+\alpha x+\frac{\alpha(\alpha-1)}{2!} x^2+\frac{\alpha(\alpha-1)(\alpha-2)}{3!} x^3+\dots +\binom{\alpha}{n}x^n+\dotsm,$$
where $\binom{\alpha}{n}$ is the generalised binomial coefficient:
$$\binom{\alpha}{n}=\frac{\alpha(\alpha-1)\dots(\alpha -n+1)}{n!}.$$
This binomial series converges for all $|x|<1$.
For $\sqrt{1+x}=(1+x)^{1/2}$, it begins with
$$\sqrt{1+x}=1+\frac x2 -\frac18x^2+\frac1{16}x^3-\frac5{128} x^4+\dotsm$$
A: $$
\begin{align}
\sqrt{1+x}
&=\sum_{n=0}^\infty\binom{1/2}{n}x^n\\
&=1+\sum_{n=0}^\infty\frac{(-1)^n\binom{2n}{n}\,x^{n+1}}{(n+1)\,2^{2n+1}}\\
&=1+\sum_{n=0}^\infty\frac{(-1)^n}{2^{2n+1}}\,C_n\,x^{n+1}
\end{align}
$$
Where $C_n$ are the Catalan Numbers.
