Answer:
The Taylor series is [tex]\ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}[/tex].
The radius of convergence is [tex]R=3[/tex].
Step-by-step explanation:
The Taylor expansion.
Recall that as we want the Taylor series centered at [tex]a=3[/tex] its expression is given in powers of [tex](x-3)[/tex]. With this in mind we need to do some transformations with the goal to obtain the asked Taylor series from the Taylor expansion of [tex]\ln(1+x)[/tex].
Then,
[tex]\ln(x) = \ln(x-3+3) = \ln(3(\frac{x-3}{3} + 1 )) = \ln 3 + \ln(1 + \frac{x-3}{3}).[/tex]
Now, in order to make a more compact notation write [tex]\frac{x-3}{3}=y[/tex]. Thus, the above expression becomes
[tex]\ln(x) = \ln 3 + \ln(1+y).[/tex]
Notice that, if x is very close from 3, then y is very close from 0. Then, we can use the Taylor expansion of the logarithm. Hence,
[tex]\ln(x) = \ln 3 + \ln(1+y) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{y^n}{n}.[/tex]
Now, substitute [tex]\frac{x-3}{3}=y[/tex] in the previous equality. Thus,
[tex]\ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}.[/tex]
Radius of convergence.
We find the radius of convergence with the Cauchy-Hadamard formula:
[tex]R^{-1} = \lim_{n\rightarrow\infty} \sqrt[n]{|a_n|},[/tex]
Where [tex]a_n[/tex] stands for the coefficients of the Taylor series and [tex]R[/tex] for the radius of convergence.
In this case the coefficients of the Taylor series are
[tex] a_n = \frac{(-1)^{n+1}}{ n3^n}[/tex]
and in consequence [tex]|a_n| = \frac{1}{3^nn}[/tex]. Then,
[tex] \sqrt[n]{|a_n|} = \sqrt[n]{\frac{1}{3^nn}}[/tex]
Applying the properties of roots
[tex] \sqrt[n]{|a_n|} = \frac{1}{3\sqrt[n]{n}}.[/tex]
Hence,
[tex]R^{-1} = \lim_{n\rightarrow\infty} \frac{1}{3\sqrt[n]{n}} =\frac{1}{3}[/tex]
Recall that
[tex]\lim_{n\rightarrow\infty} \sqrt[n]{n}=1.[/tex]
So, as [tex]R^{-1}=\frac{1}{3}[/tex] we get that [tex]R=3[/tex].
