Chapter 8.7, Problem 4E

To determine

To find: Taylor series for and centred at $4$ if $f^{\left(n\right)}\left(4\right)=\frac{{\left(-1\right)}^{n}n!}{3^n\left(n+1\right)}$ and it’s radius and convergence.

Answer to Problem 4E

The Taylor series expansion of $f$ at centre $4$ is ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n}{3^n\left(n+1\right)}\cdot {\left(x-4\right)}^n}$ and Radius of convergence is $3.$

Explanation of Solution

Given information: $f^{\left(n\right)}\left(4\right)=\frac{{\left(-1\right)}^{n}n!}{3^n\left(n+1\right)}$

Calculation:

We know that the Taylor series expansion of $f$ at $x$ is ${\displaystyle \sum _{n=0}^{\infty }\frac{f^{\left(n\right)}\left(x_0\right)}{n!}{\left(x-x_0\right)}^n}$

Hence, the Taylor series expansion for $f$ centred at $4$ is

  $\begin{array}{l}{\displaystyle \sum _{n=0}^{\infty }\frac{\frac{{\left(-1\right)}^{n}n!}{3^n\left(n+1\right)}}{n!}\cdot {\left(x-4\right)}^n}\\ ={\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n}{3^n\left(n+1\right)}\cdot {\left(x-4\right)}^n}\end{array}$

To find the Radius and Convergence:-

Here, $a_n=\frac{{\left(-1\right)}^n}{3^n\left(n+1\right)}$

  $\begin{array}{l}\frac{1}{R}=\underset{n\to \infty }{\lim }\left|\frac{a_n+1}{a_n}\right|\\ \Rightarrow \frac{1}{R}=\underset{n\to \infty }{\lim }\left|\frac{{\left(-1\right)}^{n+1}}{3^{n+1}\left(n+2\right)}\cdot \frac{3^n\left(n+1\right)}{{\left(-1\right)}^n}\right|\\ \Rightarrow \frac{1}{R}=\underset{n\to \infty }{\lim }\frac{\left(n+1\right)}{\left(n+2\right)}\cdot \frac{1}{3}\\ \Rightarrow \frac{1}{R}=\underset{n\to \infty }{\lim }\frac{\left(1+\frac{1}{n}\right)}{\left(1+\frac{2}{n}\right)}\cdot \frac{1}{3}\\ \Rightarrow \frac{1}{R}=\frac{1}{3}\\ i.e.,R=3\end{array}$

Hence, Radius of convergence is $3$ and Interval of convergence is $\left(-3+4,3+4\right)$ i.e. $\left(1,7\right)$

[Since, Centre $=4$ ]