Taylor series and interval of convergence a. Use the definition of a Taylor/Maclaurin series to...

A) The Taylor series for a function f(x) continuously differentiable at $x=a$ is given by,
$f(x)=f(a)+\frac{(x-a)}{1!}{f}^{\prime }(a)+\frac{(x-a{)}^2}{2!}{f}^{″}(a)+\frac{(x-a{)}^3}{3!}{f}^{‴}(a)...$
Here
$f(x)={\log }_3(x+1)$
$a=0,$
Substitute the values,
$f(x)=f(a)+\frac{(x-a)}{1!}{f}^{\prime }(a)+\frac{(x-a{)}^2}{2!}{f}^{″}(a)+\frac{(x-a{)}^3}{3!}{f}^{‴}(a)...$
${\log }_3(x+1)={\log }_3(x+1)+\frac{x}{1!}(\frac{1}{(x+1)\log 3})+\frac{x^2}{2\times 1}(\frac{-1}{(x+1{)}^2\log 3})+\frac{x^3}{3\times 2\times 1}[-(\frac{-2}{(x+1{)}^3\log 3})]...$
$=\frac{\log (x+1)}{\log (3)}+\frac{x}{(x+1)\log 3}-\frac{x^2}{2(x+1{)}^2\log 3}+\frac{x^3}{3(x+1{)}^3\log 3}...$
The first four terms of the Taylor's expansion of $f(x)={\log }_3(x+1)$ are respectively,
$\frac{\log (x+1)}{{\log }_3},\frac{x}{(x+1)\log 3},\frac{x^2}{2(x+1{)}^2\log 3},\frac{x^3}{3(x+1{)}^3\log 3}$
B) The power series for the above Taylor's expansion is given by,
${\log }_3(x+1)=\sum _{n=0}^{\mathrm{\infty }}\{\frac{(x-a{)}^n}{n!}\cdot f^n(a)\}$
Re-arrange the series,
${\log }_3(x+1)=\frac{x}{(x+1)\log (3)}+\frac{x}{(x+1)\log (3)}-\frac{1}{\log (3)}[\frac{x^2}{2(x+1{)}^2}-\frac{x^3}{3(x+1{)}^3}+\frac{x^4}{4(x+1{)}^4}-\frac{x^5}{5(x+1{)}^5}+\frac{x^6}{6(x+1{)}^6}...]$
$=\frac{x}{(x+1)\log (3)}+\frac{x}{(x+1)\log (3)}-\frac{1}{\log (3)}\sum _{n=2}^{\mathrm{\infty }}[\frac{x^n}{n(x+1{)}^n}\cdot (-1{)}^n]$
The series will be converging if the term $\sum _{n=2}^{\mathrm{\infty }}[\frac{x^n}{n(x+1{)}^n}\cdot (-1{)}^n]$ is converging,
Let