Find the Taylor series centered at zero for the function f(x)=ln(2 + x^2). Determine the radius of convergence of this series.

Find the Taylor series centered at zero for the function $f(x)=\ln (2+x^2)$. Determine the radius of convergence of this series.
Consider a given function is
$f(x)=\ln (2+x^2)$.
Now,
$f(x)=\ln [2(1+\frac{x^2}{2})]$
$=\ln (2)+\ln (1+\frac{x^2}{2})$
$=\ln (2)+\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}(\frac{x^2}{2}{)}^n}{n}...[\ln (1+x)=\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}{x}^n}{n}]$
$=\ln (2)+\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}{x}^{2n}}{n{2}^n}$
Hence, the required Taylor's series of $f(x)=\ln (2+x^2)$ is
$\ln (2)+\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}{x}^{2n}}{n{2}^n}$
Consider a Taylor's series is $\ln (2)+\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}{x}^{2n}}{n{2}^n}$
To find a Taylor's series, consider a sequence $<\frac{(-1{)}^{n+1}{x}^{2n}}{n{2}^n}>$
Now, by ratio test,
$L=\underset{n\to \mathrm{\infty }}{lim}|\frac{a_{n+1}}{a_n}|$
$=\underset{n\to \mathrm{\infty }}{lim}|\frac{\frac{(-1{)}^{n+1}{x}^{2n+2}}{(n+1)2^{n+1}}}{\frac{(-1{)}^{n+1}{x}^{2n}}{n{2}^n}}|$
$=\underset{n\to \mathrm{\infty }}{lim}|\frac{n{2}^n{x}^{2n+2}}{(n+1)2^{n+1}{x}^{2n}}|$
$=\underset{n\to \mathrm{\infty }}{lim}|\frac{n{x}^2}{(n+1)2}|$
$=|\frac{x^2}{2}|$

The series is convergent if $L<1$. Hence, $|\frac{x^2}{2}|<1$

It implies that 0

Check at the end point $x=\sqrt{2}$. Then,

$\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}{x}^{2n}}{n{2}^n}=\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}(\sqrt{2}{)}^{2n}}{n{2}^n}$

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

It is convergent series.

Hence, the radius of convergence is $0$