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:

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