Taylor series of $\sqrt{1+x}$ using sigma notation I want help in writing Taylor series of $\sqrt{1+x}$ using sigma notation
I got till 
$1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\frac{5x^4}{128}+\ldots$ and so on.
But I don't know what will come in sigma notation.
 A: If $f(x) = \sqrt{1+x} = (1+x)^\frac{1}{2}$, then $f'(x) = \frac{1}{2}(1+x)^{-\frac{1}{2}}$ and $f''(x) = \frac{1}{2}\frac{-1}{2}(1+x)^{-\frac{3}{2}}$. In general, $$
  \frac{d}{dx}(1 + x)^{\frac{-n}{2}} = \frac{-n}{2}(1 + x)^{\frac{-n-2}{2}} \text{,}
$$
and therefore the $n$-th derivative of $f(x) = \sqrt{1+x}$ is (for $n > 1$) $$
  \frac{1}{2}\underbrace{\frac{-1}{2}\frac{-3}{2}\ldots\frac{-2n+3}{2}}_{n-1\textrm{ terms}}(1+x)^{-\frac{-2n+1}{2}} \text{.}
$$
Evaluating at $x=0$ yields (again for $n > 1$) $$
  f^{(n)}(0) = (-1)^{n-1}\frac{1\cdot 3 \cdot \ldots \cdot (2n-3)}{2^n}
$$
and the taylor series around $x=0$ is thus $$
  \sum_{n=0}^\infty x^n\frac{f^{(n)}(0)}{n!} =  1 + x\frac{1}{2} - x^2\frac{1}{8} + \sum_{n=3}^\infty x^n (-1)^{n-1}\frac{1\cdot 3 \cdot \ldots \cdot (2n-3)}{n!2^n} \text{.}
$$
Cutting this off at $x^4$ yields $$
  \sqrt{1+x} \approx  1 + x\frac{1}{2} - x^2\frac{1}{8} + x^3\frac{1\cdot 3}{3!2^3} - x^4\frac{1\cdot 3\cdot 5}{4!2^4}
  = 1 + x\frac{1}{2} - x^2\frac{1}{8} + x^3\frac{1}{16} - x^4\frac{5}{128}
$$
so your initial coefficients are correct.
You can write the series in terms of factorials by using that $1\cdot 3 \cdots\ldots\cdot (2k+1) = \frac{(2k+1)!}{2 \cdot 4\cdot\ldots\cdot (2k)}$ and that $2 \cdot 4\cdot\ldots\cdot (2k) = (2\cdot1)\cdot(2\cdot 2)\cdot(2\cdot 3)\cdot\ldots\cdot(2k) = k!2^k$. Overall, $$
  1\cdot 3 \cdots\ldots\cdot (2k+1) = \frac{(2k+1)!}{k!2^k}
$$
and therefore ($k = n - 2$) $$
  \frac{1\cdot 3 \cdot \ldots \cdot (2n-3)}{n!2^n}
  = \frac{(2n-3)!}{n!2^n(n-2)!2^{n-2}}
  = \frac{(2n-3)!}{n!(n-2)!2^{2n-2}} \text{,}
$$
so the taylor series becomes $$
  \sum_{n=0}^\infty x^n\frac{f^{(n)}(0)}{n!} =  1 + x\frac{1}{2} - x^2\frac{1}{8} + \sum_{n=3}^\infty x^n (-1)^{n-1}\frac{(2n-3)!}{n!(n-2)!2^{2n-2}} \text{.}
$$
A: The generalized binomial theorem says that
$$
(1+x)^{1/2}=\sum_{k=0}^\infty\binom{1/2}{k}x^k
$$
where $\binom{1/2}{0}=1$ and for $k\ge1$,
$$
\begin{align}
\binom{1/2}{k}
&=\frac{\frac12(\frac12-1)(\frac12-2)\cdots(\frac12-k+1)}{k!}\\
&=\frac{(-1)^{k-1}}{2^kk!}1\cdot3\cdot5\cdots(2k-3)\\
&=\frac{(-1)^{k-1}}{2^kk!}\frac{(2k-2)!}{2^{k-1}(k-1)!}\\
&=\frac{(-1)^{k-1}}{k2^{2k-1}}\binom{2k-2}{k-1}
\end{align}
$$
Thus,
$$
\begin{align}
(1+x)^{1/2}
&=1-\sum_{k=1}^\infty\frac2k\binom{2k-2}{k-1}\left(-\frac x4\right)^k\\
&=1-\sum_{k=0}^\infty\frac2{k+1}\binom{2k}{k}\left(-\frac x4\right)^{k+1}
\end{align}
$$
A: allow me to take one more tiny little step forward based on your excellent calculations.
$$
\sqrt{1+x}=\sum _{n=0}^{\infty } \frac{ 1}{(1-2 n)}
\frac{(2 n)!}{n! n!}\left(-\frac{x}{4}\right)^n
$$
or
$$
\sqrt{1+x}=\sum _{n=0}^{\infty } \frac{ C_{2n}^n}{(1-2 n)}
\left(-\frac{x}{4}\right)^n
$$
