Taylor expansion of $\sqrt{1+x}$

The Taylor series for  $f(x)=\sqrt{1+x}$  using the

$$T(x)=\sum _{k=0}^{\mathrm{\infty }}\frac{f^{(k)}(a)}{k!}{(x-a)}^k$$

is given in the table below for the first few .

$k$	expansion	simplified	at  $a=0$
0	$f(a)$	${(1+a)}^{1/2}$	$1$
1	$f^{\prime }(a)(x-a)$	${\displaystyle \frac{1}{2}}{(1+a)}^{-1/2}(x-a)$	${\displaystyle \frac{1}{2}}x$
2	${\displaystyle \frac{f^{(2)}(a)}{2!}}{(x-a)}^2$	$-{\displaystyle \frac{1}{8}}{(1+a)}^{-3/2}{(x-a)}^2$	$-{\displaystyle \frac{1}{8}}{x}^2$
3	${\displaystyle \frac{f^{(3)}(a)}{3!}}{(x-a)}^3$	${\displaystyle \frac{3}{48}}{(1+a)}^{-5/2}{(x-a)}^3$	${\displaystyle \frac{1}{16}}{x}^3$
4	${\displaystyle \frac{f^{(4)}(a)}{4!}}{(x-a)}^4$	$-{\displaystyle \frac{15}{384}}{(1+a)}^{-7/2}{(x-a)}^4$	$-{\displaystyle \frac{5}{128}}{x}^4$
5	${\displaystyle \frac{f^{(5)}(a)}{5!}}{(x-a)}^5$	${\displaystyle \frac{105}{3840}}{(1+a)}^{-9/2}{(x-a)}^5$	${\displaystyle \frac{7}{256}}{x}^5$
6	${\displaystyle \frac{f^{(6)}(a)}{6!}}{(x-a)}^6$	$-{\displaystyle \frac{945}{46080}}{(1+a)}^{-11/2}{(x-a)}^6$	$-{\displaystyle \frac{21}{1024}}{x}^6$

Table 1: Taylor Series for $f(x)=\sqrt{1+x}$

The general coefficient of the expansion, for $n\ge 2$is:

${\displaystyle \frac{f^{(n)}(a)}{n!}}$$=\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{1}{2}}{n}}\right){(1+a)}^{\frac{1}{2}-n}$

$={\displaystyle \frac{\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\mathrm{\cdots }\left(\frac{1}{2}-(n-1)\right)}{n(n-1)(n-2)\mathrm{\cdots }1}}\mathit{\hspace{1em}}{(1+a)}^{-\frac{2n-1}{2}}$

$={\displaystyle \frac{\frac{1}{2}{(-\frac{1}{2})}^{n-1}\mathit{\hspace{1em}}1\cdot 3\mathrm{\cdots }(2n-3)}{n(n-1)(n-2)\mathrm{\cdots }1}}\mathit{\hspace{1em}}{(1+a)}^{-\frac{2n-1}{2}}$

$={\displaystyle \frac{{(-1)}^{n-1}}{2^{n}n!}}{\displaystyle \frac{(2n-3)!}{(2n-4)(2n-6)\mathrm{\cdots }2}}\mathit{\hspace{1em}}{(1+a)}^{-\frac{2n-1}{2}}$

$={(-1)}^{n-1}{\displaystyle \frac{(2n-3)!}{2^{2n-2}n!(n-2)!}}\mathit{\hspace{1em}}{(1+a)}^{-\frac{2n-1}{2}}.$