A) The Taylor series for a function f(x) continuously differentiable at \(x=a\) is given by,

\(f(x)=f(a)+\frac{(x-a)}{1!}f'(a)+\frac{(x-a)^2}{2!}f''(a)+\frac{(x-a)^3}{3!}f'''(a)...\)

Here

\(f(x)=\log_3(x+1)\)

\(a=0,\)

Substitute the values,

\(f(x)=f(a)+\frac{(x-a)}{1!}f'(a)+\frac{(x-a)^2}{2!}f''(a)+\frac{(x-a)^3}{3!}f'''(a)...\)

\(\log_3(x+1)=\log_3(x+1)+\frac{x}{1!}(\frac{1}{(x+1)\log3})+\frac{x^2}{2\times1}(\frac{-1}{(x+1)^2\log3})+\frac{x^3}{3\times2\times1}[-(\frac{-2}{(x+1)^3\log3})]...\)

\(=\frac{\log(x+1)}{\log(3)}+\frac{x}{(x+1)\log3}-\frac{x^2}{2(x+1)^2\log3}+\frac{x^3}{3(x+1)^3\log3}...\)

The first four terms of the Taylor's expansion of \(f(x)=\log_3(x+1)\) are respectively,

\(\frac{\log(x+1)}{\log_3},\frac{x}{(x+1)\log3},\frac{x^2}{2(x+1)^2\log3},\frac{x^3}{3(x+1)^3\log3}\)

B) The power series for the above Taylor's expansion is given by,

\(\log_3(x+1)=\sum_{n=0}^\infty\left\{\frac{(x-a)^n}{n!}\cdot f^n(a)\right\}\)

Re-arrange the series,

\(\log_3(x+1)=\frac{x}{(x+1)\log(3)}+\frac{x}{(x+1)\log(3)}-\frac{1}{\log(3)}[\frac{x^2}{2(x+1)^2}-\frac{x^3}{3(x+1)^3}+\frac{x^4}{4(x+1)^4}-\frac{x^5}{5(x+1)^5}+\frac{x^6}{6(x+1)^6}...]\)

\(=\frac{x}{(x+1)\log(3)}+\frac{x}{(x+1)\log(3)}-\frac{1}{\log(3)}\sum_{n=2}^\infty[\frac{x^n}{n(x+1)^n}\cdot(-1)^n]\)

The series will be converging if the term \(\sum_{n=2}^\infty[\frac{x^n}{n(x+1)^n}\cdot(-1)^n]\) is converging,

Let \(a_n=(-1)^n\cdot\frac{x^n}{n(x+1)^n}\)

Apply the Ratio Test,

\(L=\lim_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|\)

\(=\lim_{n\rightarrow\infty}|\frac{[\frac{x^{n+1}}{(n+1)(x+1)^{n+1}}\cdot(-1)^{n+1}]}{[\frac{x^n}{n(x+1)^n}\cdot(-1)^n]}|\)

\(=\lim_{n\rightarrow\infty}|\frac{-xn}{(x+1)(n+1)}|\)

\(=|\frac{x}{x+1}|\cdot\lim_{n\rightarrow\infty}|\frac{1}{1+\frac1n}|\)

\(=|\frac{x}{x+1}|\)

The series is converging if,

\( L<1\)

\(|\frac{x}{x+1}|<1\)

\(-1<\frac{x}{x+1}<1\)

The term \(\frac{x}{x+1}<1\) is true for any real value of \(x \in R,\)

Also,

\(\frac{x}{x+1}<1\)

\(x>-x-1\)

\(2x>1\)

\(x>\frac{-1}{2}\)

Thus, the interval of convergence for the given power series is \((\frac{-1}{2},\infty)\)