Answer
(a) Please see the figure attached.
The graphs suggest that the interval of convergence of the Taylor series for $f$ is $ - 1 \lt x \lt 1$.
(b) The Taylor expansion for $f$ is valid at $x=1$ and at $x=-1$.
Hence, the interval of convergence of the Taylor series for $f\left( x \right) = \sqrt {1 + x} $ is $ - 1 \le x \le 1$.
Work Step by Step
(a) Write $f\left( x \right) = \sqrt {1 + x} = {\left( {1 + x} \right)^{1/2}}$.
By Theorem 3:
${\left( {1 + x} \right)^a} = 1 + \dfrac{a}{{1!}}x + \dfrac{{a\left( {a - 1} \right)}}{{2!}}{x^2} + \dfrac{{a\left( {a - 1} \right)\left( {a - 2} \right)}}{{3!}}{x^3} + \cdot\cdot\cdot + \left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right){x^n} + \cdot\cdot\cdot = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right){x^n}$
converges for $\left| x \right| \lt 1$.
Thus,
$f\left( x \right) = \sqrt {1 + x} = {\left( {1 + x} \right)^{1/2}} = 1 + \dfrac{1}{2}x + \dfrac{{\left( {\dfrac{1}{2}} \right)\left( { - \dfrac{1}{2}} \right)}}{{2!}}{x^2} + \dfrac{{\left( {\dfrac{1}{2}} \right)\left( { - \dfrac{1}{2}} \right)\left( { - \dfrac{3}{2}} \right)}}{{3!}}{x^3} + \dfrac{{\left( {\dfrac{1}{2}} \right)\left( { - \dfrac{1}{2}} \right)\left( { - \dfrac{3}{2}} \right)\left( { - \dfrac{5}{2}} \right)}}{{4!}}{x^4} + \cdot\cdot\cdot = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{1/2}\\
n
\end{array}} \right){x^n}$
We list in the table below the the first five Taylor polynomials for $f$:
$\begin{array}{*{20}{c}}
N&{}&{\mathop \sum \limits_{n = 0}^N \left( {\begin{array}{*{20}{c}}
{1/2}\\
n
\end{array}} \right){x^n}}\\
0&{}&1\\
1&{}&{1 + \dfrac{x}{2}}\\
2&{}&{1 + \dfrac{x}{2} - \dfrac{{{x^2}}}{8}}\\
3&{}&{1 + \dfrac{x}{2} - \dfrac{{{x^2}}}{8} + \dfrac{{{x^3}}}{{16}}}\\
4&{}&{1 + \dfrac{x}{2} - \dfrac{{{x^2}}}{8} + \dfrac{{{x^3}}}{{16}} - \dfrac{{5{x^4}}}{{128}}}
\end{array}$
Using a computer algebra system we plot the graph of $f\left( x \right) = \sqrt {1 + x} $ (the red curve) with the graphs of the first five Taylor polynomials for $f$ (the dashed curves), as is shown in the figure attached.
The graphs suggest that the interval of convergence of the Taylor series for $f$ is $ - 1 \lt x \lt 1$. In part (b) below, we investigate the endpoints: $x=1$ and $x=-1$, numerically.
(b) We compute some values of the Taylor polynomials for $f$ and list them in the following tables:
1. For $x=1$
$\begin{array}{*{20}{c}}
N&{}&{\mathop \sum \limits_{n = 0}^N \left( {\begin{array}{*{20}{c}}
{1/2}\\
n
\end{array}} \right)}\\
{10}&{}&{1.40993}\\
{20}&{}&{1.41267}\\
{50}&{}&{1.41382}\\
{100}&{}&{1.41407}\\
{1000}&{}&{1.41421}
\end{array}$
The results show that the series converges to $1.41421$ as $N \to \infty $. Since $f\left( 1 \right) = \sqrt 2 \approx 1.4142$, we conclude that the Taylor expansion for $f$ is valid at $x=1$.
2. For $x=-1$
$\begin{array}{*{20}{c}}
N&{}&{\mathop \sum \limits_{n = 0}^N \left( {\begin{array}{*{20}{c}}
{1/2}\\
n
\end{array}} \right){{\left( { - 1} \right)}^n}}\\
{10}&{}&{0.176197}\\
{50}&{}&{0.0795892}\\
{100}&{}&{0.0563485}\\
{1000}&{}&{0.017839}\\
{10000}&{}&{0.00564183}
\end{array}$
The results show that the series converges to $0$ as $N \to \infty $. Since $f\left( { - 1} \right) = 0$, we conclude that the Taylor expansion for $f$ is valid at $x=-1$.
Hence, the interval of convergence of the Taylor series for $f\left( x \right) = \sqrt {1 + x} $ is $ - 1 \le x \le 1$.
