## Thomas' Calculus 13th Edition

$1+ \dfrac{1}{2}x-\dfrac{1}{8}x^2+\dfrac{1}{16}x^3-....+(-1)^{n-1} \dfrac{1 \cdot 3 \cdot 5...(2n-3)x^n }{n! 2^n}$
Differentiate the given function $f(x)$ as follows: $f'(x)=\dfrac{1}{2 (x+1)^{1/2}} \implies f'(0)=1$; $f''(x)=-\dfrac{1}{4(x+1)^{3/2}} \implies f''(0)=\dfrac{1}{2} \\ f'''(x)=\dfrac{3}{8(x+1)^{5/2}}\implies f'''(0)=-\dfrac{1}{4}$ The Maclaurin series at $x=0$ can be written as: $f(x)=\sqrt {x+1}=1+ \dfrac{1}{2}x-\dfrac{1}{8}x^2+\dfrac{1}{16}x^3-....+(-1)^{n-1} \dfrac{1 \cdot 3 \cdot 5...(2n-3) x^n}{n! 2^n}$