Question

Use the series you know to show that:

$\mathit{l}\mathit{n}\mathbf{3}\mathbf{+}\frac{{\mathbf{\left(}\mathbf{l}\mathbf{n}\mathbf{3}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{2}\mathbf{!}}\mathbf{+}\frac{{\mathbf{\left(}\mathbf{l}\mathbf{n}\mathbf{3}\mathbf{\right)}}^{\mathbf{3}}}{\mathbf{3}\mathbf{!}}\mathbf{+}\mathbf{\dots }\mathbf{=}\mathbf{2}$

Short answer

It is shown that the expansion of series$\ln \left(3\right)+\frac{\ln {\left(3\right)}^2}{2!}+\frac{\ln {\left(3\right)}^3}{3!}+\frac{\ln {\left(3\right)}^4}{4!}+\dots \text{is}2$

Step-by-step solution

Maclaurin series and the given series.

Series is$\ln \left(3\right)+\frac{\ln {\left(3\right)}^2}{2!}+\frac{\ln {\left(3\right)}^3}{3!}+\frac{\ln {\left(3\right)}^4}{4!}+\dots$
The Maclaurin series ofis given below:

$$\begin{gathered} \begin{array}{rcl}& & {\mathbf{e}}^{\mathbf{x}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}\mathbf{!}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}\mathbf{!}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{4}\mathbf{!}}\mathbf{+}\mathbf{\dots } \\ {\mathbf{e}}^{\mathbf{x}}\mathbf{-}\mathbf{1}\mathbf{=}\mathbf{x}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}\mathbf{!}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}\mathbf{!}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{4}\mathbf{!}}\mathbf{+}\mathbf{\dots }\\ & & \end{array} \end{gathered}$$

Use the Maclaurin series.

Substitute$\ln \left(3\right)$ for$x$ in the Maclaurin series of $e^x-1=x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\dots$as follows:

$e^{\ln \left(3\right)}-1=\ln \left(3\right)+\frac{\ln {\left(3\right)}^2}{2!}+\frac{\ln {\left(3\right)}^3}{3!}+\frac{\ln {\left(3\right)}^4}{4!}+\dots$

The value of $e^{\ln \left(3\right)}$is given below:

$e^{\ln \left(3\right)}=3$

Substitute $3$for$e^{\ln \left(3\right)}$ in the equation $e^{\ln \left(3\right)}-1=\ln \left(3\right)+\frac{\ln {\left(3\right)}^2}{2!}+\frac{\ln {\left(3\right)}^3}{3!}+\frac{\ln {\left(3\right)}^4}{4!}+\dots$as follows:

$$\begin{gathered} 3-1=\ln \left(3\right)+\frac{\ln {\left(3\right)}^2}{2!}+\frac{\ln {\left(3\right)}^3}{3!}+\frac{\ln {\left(3\right)}^4}{4!}+\dots \\ 2=\ln \left(3\right)+\frac{\ln {\left(3\right)}^2}{2!}+\frac{\ln {\left(3\right)}^3}{3!}+\frac{\ln {\left(3\right)}^4}{4!}+\dots \end{gathered}$$

Thus, it is shown that the expansion of series$\ln \left(3\right)+\frac{\ln {\left(3\right)}^2}{2!}+\frac{\ln {\left(3\right)}^3}{3!}+\frac{\ln {\left(3\right)}^4}{4!}+\dots \text{is}2$