Answer
\[\boxed{\begin{array}{*{20}{c}}
n&{{x_n}} \\
0&{1.700000} \\
1&{1.718220} \\
2&{1.718282} \\
3&{1.718282} \\
4&{1.718282} \\
5&{1.718282} \\
6&{1.718282} \\
7&{1.718282} \\
8&{1.718282} \\
9&{1.718282} \\
{10}&{1.718282}
\end{array}}\]
Work Step by Step
\[\begin{gathered}
{\text{Let }}f\left( x \right) = \ln \left( {x + 1} \right) - 1,{\text{ and }}{x_0} = 1.7 \hfill \\
{\text{Using the Newton's Method }}{x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} \hfill \\
f'\left( x \right) = \frac{1}{{x + 1}} \hfill \\
{\text{Then evaluating }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right) \hfill \\
{x_{n + 1}} = {x_n} - \frac{{\ln \left( {{x_n} + 1} \right) - 1}}{{\frac{1}{{{x_n} + 1}}}} \hfill \\
\hfill \\
{\text{Taking }}{x_0} = 1.7 \hfill \\
{x_0} = 1.7 \hfill \\
{x_{0 + 1}} = {x_1} = 1.7 - \frac{{\ln \left( {1.7 + 1} \right) - 1}}{{\frac{1}{{1.7 + 1}}}} = 1.718220 \hfill \\
{x_{1 + 1}} = {x_2} = 1.718220 - \frac{{\ln \left( {1.718220 + 1} \right) - 1}}{{\frac{1}{{1.718220 + 1}}}} \approx 1.718282 \hfill \\
{x_{1 + 1}} = {x_2} = 1.718282 - \frac{{\ln \left( {1.718282 + 1} \right) - 1}}{{\frac{1}{{1.718282 + 1}}}} \approx 1.718282 \hfill \\
{x_{1 + 3}} = {x_4} = 1.718282 - \frac{{\ln \left( {1.718282 + 1} \right) - 1}}{{\frac{1}{{1.718282 + 1}}}} \approx 1.718282 \hfill \\
{x_{1 + 4}} = {x_5} = 1.718282 - \frac{{\ln \left( {1.718282 + 1} \right) - 1}}{{\frac{1}{{1.718282 + 1}}}} \approx 1.718282 \hfill \\
{x_{1 + 5}} = {x_6} = 1.718282 - \frac{{\ln \left( {1.718282 + 1} \right) - 1}}{{\frac{1}{{1.718282 + 1}}}} \approx 1.718282 \hfill \\
{x_{1 + 6}} = {x_7} = 1.718282 - \frac{{\ln \left( {1.718282 + 1} \right) - 1}}{{\frac{1}{{1.718282 + 1}}}} \approx 1.718282 \hfill \\
{x_{1 + 7}} = {x_8} = 1.718282 - \frac{{\ln \left( {1.718282 + 1} \right) - 1}}{{\frac{1}{{1.718282 + 1}}}} \approx 1.718282 \hfill \\
{x_{1 + 8}} = {x_9} = 1.718282 - \frac{{\ln \left( {1.718282 + 1} \right) - 1}}{{\frac{1}{{1.718282 + 1}}}} \approx 1.718282 \hfill \\
{x_{1 + 9}} = {x_{10}} = 1.718282 - \frac{{\ln \left( {1.718282 + 1} \right) - 1}}{{\frac{1}{{1.718282 + 1}}}} \approx 1.718282 \hfill \\
\hfill \\
{\text{Thus}}{\text{, we obtain}} \hfill \\
\boxed{\begin{array}{*{20}{c}}
n&{{x_n}} \\
0&{1.700000} \\
1&{1.718220} \\
2&{1.718282} \\
3&{1.718282} \\
4&{1.718282} \\
5&{1.718282} \\
6&{1.718282} \\
7&{1.718282} \\
8&{1.718282} \\
9&{1.718282} \\
{10}&{1.718282}
\end{array}} \hfill \\
\end{gathered} \]
