Chapter 10, Problem 18RE

To determine

To find: The function $f\left(x\right)$ , value of x and the sum of the given series.

Answer to Problem 18RE

Function is $n\left(1+x\right)$ , point is $x=\frac{2}{3}$ and sum is $S=\ln \frac{5}{3}$

Explanation of Solution

Given:

The given Maclaurin series for the function $f\left(x\right)$ at a particular point is

  $S=\frac{2}{3}-\frac{4}{18}+\frac{8}{81}-\mathrm{....}+{\left(-1\right)}^{n-1}\frac{2^n}{n{3}^n}+,,,$

Calculation:

As, the Maclaurin series for the function $f\left(x\right)=\ln \left(1+x\right)$ is

  $f\left(x\right)=\ln \left(1+x\right)=x-\frac{x^2}{2}+\frac{x^3}{3}-\mathrm{....}={\left(-1\right)}^{n-1}\frac{x^n}{n}+\mathrm{....}$

Now, at $x=\frac{2}{3}$

  $f\left(\frac{2}{3}\right)=\frac{2}{3}-\frac{4}{18}+\frac{8}{81}-\mathrm{....}+{\left(-1\right)}^{n-1}\frac{2^n}{n{3}^n}+,,,$

This series is similar to the given series.

So, the function is

  $f\left(x\right)=\ln \left(1+x\right)$ and $x=\frac{2}{3}$

And the sum of the given series is,

  $\begin{array}{l}f\left(\frac{2}{3}\right)=\ln \left(1+\frac{2}{3}\right)\\ \Rightarrow S=\ln \frac{5}{3}\end{array}$

Therefore, function is $\ln \left(1+x\right)$ , point is $x=\frac{2}{3}$ and sum is $S=\ln \frac{5}{3}$