Question: The Taylor series for f(x) =ln (x)about x₂=0given in equation (13) can also be obtained as follows:

(a)Starting with the expansion 1/ (1-s) =$\sum _{n=0}^{\infty }s\text{'}\text{'}$ and observing that

'

obtain the Taylor series for 1/xabout x₀= 1.

(b)Since use the result of part (a) and termwise integration to obtain the Taylor series for f (x)=lnxaboutx₀= 1.

For a function^{$f\left(x\right)$} the Taylor series expansion about a point ^x₀ is given by,$f(x-x_0)=f\left(x_0\right)+f\text{'}\left(x_0\right).(x-x_0)+f\text{'}\text{'}\left(x_0\right).\frac{{(x-x_0)}^2}{2!}+f\text{'}\text{'}\text{'}\left(x_0\right).\frac{{(x-x_0)}^3}{3!}+...$

Rewrite $\frac{1}{x}$by adding and subtracting 1 from the denominator, as follows:

$\frac{1}{x}=\frac{1}{1+(x-1)}$and$\frac{1}{1-s}=\sum _{n=0}^{\infty }s\text{'}\text{'}$

It follows,

$$\begin{gathered} \frac{1}{x}=\frac{1}{1+(x-1)} \\ =\sum _{n=1}^{\infty }{(1-x)}^n \\ =\sum _{n=1}^{\infty }{{(-1)}^n(1-x)}^n \end{gathered}$$

Therefore, the Taylor series for$\frac{1}{x}$ around x₀=1 is given by: .$\frac{1}{x}$role="math" localid="1664187564177" $=\sum _{n=1}^{\infty }{(-1)}^n(1-x)$

Substitute $\frac{1}{x}=\sum _{n=1}^{\infty }{(-1)}^n{(1-x)}^n$in the equation role="math" localid="1664187704795" $\ln x={\int }_1^x\frac{1}{t}dt,$we get,

$\ln x={\int }_1^x\sum _{n=1}^{\infty }{(-1)}^n{(t-1)}^{n}dt$

Interchange the integral and the sum,

role="math" localid="1664188068704" $\ln x=\sum _{n=1}^{\infty }{\int }_1^x{(-1)}^n{(t-1)}^{n}dt$

$\ln x=\sum _{n=1}^{\infty }{(-1)}^n{\int }_1^x{(t-1)}^{n}dt$

Take integration,

$\ln x=\sum _{n=1}^{\infty }{(-1)}^n{{\overline{)\frac{{(t-1)}^{n+1}}{n+1}}}^x}_1$

Compute the boundaries,

$\ln x=\sum _{n=1}^{\infty }{(-1)}^n\frac{{(t-1)}^{n+1}}{n+1}$

Let,

k = n + 1

n = k - 1

So,

$\ln x=\sum _{\mathrm{k}=1}^{\infty }{(-1)}^{\mathrm{k}-1}\frac{{(\mathrm{x}-1)}^k}{k}$

Now substitute k for n in the above equation, it is only a dummy index the name is not important, we need to respect the range from 1 to$\infty$.