Functions to power series Find power series representations centered at 0 for the following functions using known power series. Give the interval of convergence for the resulting series. f(x)=lnsqrt{4-x}

Given, $f(x)=\ln \sqrt{4-x}$
Find power series representation, the radius of converges, the interval of converges.
Use the power series formula,
 $f(x)=\ln (1-x^n)=-\sum _{k=1}^{\mathrm{\infty }}\frac{x^k}{k},-1\le x<1$

Consider
$f(x)=\ln \sqrt{4-x^2}$
$f(x)=\ln (4-x^2{)}^{\frac{1}{2}}$
Use the power rule of logarithmic,
$\ln m^2=n\ln m$
$=\frac{1}{2}\ln (4-x^2)$
Now multiple and divide "4" on the logarithmic function we get,
$=\frac{1}{2}\ln 4(\frac{4-x^2}{4})$
$=\frac{1}{2}\ln 4(\frac{4}{4}-\frac{x^2}{4})$
$=\frac{1}{2}\ln 4(1-\frac{x^2}{4})$
Now use the product rule of logarithmic we get,
$\ln (mn)=\ln (m)+\ln (n)$
$=\frac{1}{2}\ln 4+\ln (1-\frac{x^2}{4})$
Now again use the power rule of logarithmic we get,
$=\ln 4^{\frac{1}{2}}+\ln (1-\frac{x^2}{4})$
$=\ln 2^{\frac{2}{2}}+\ln (1-\frac{x^2}{4})$
$=\ln 2+\ln (1-\frac{x^2}{4})$
Now use the power series formula we get,
$=\ln 2-\sum _{k=1}^{\mathrm{\infty }}\frac{(\frac{x^2}{4}{)}^k}{k}$
$=\ln 2-\sum _{k=1}^{\mathrm{\infty }}\frac{(\frac{x^{2k}}{4^k})}{k}$
For this we get interval, $-2\le x\le 2$
Use Ratio test to check the series converges or diverges:
Ratio test definition:
Suppose we have series $\sum a_n$ defined,
$L=\underset{n\to \mathrm{\infty }}{lim}|\frac{a_{n+1}}{a_n}|$
If $L>1$ then the series diverges,
If $L<1$ then the series converges,
If $L=1$ then the test fails.
Consider,
$a_k=\frac{x^{2k}}{4^{k}k}$
$a_{k+1}=\frac{x^{2(k+1)}}{4^{(k+1)}(k+1)}$
$\underset{k\to \mathrm{\infty }}{lim}|\frac{\frac{x^{2k}{x}^2}{4^{k}4(k+1)}}{\frac{x^{2k}}{4^{k}k}}|=\underset{k\to \mathrm{\infty }}{lim}|\frac{x^{2k}{x}^2}{4^{k}4(k+1)}\times \frac{4^{k}k}{x^{2k}}|$
$=\underset{k\to \mathrm{\infty }}{lim}|\frac{x^2}{4}|\frac{k}{k+1}|$
$=|\frac{x^2}{4}|\underset{k\to \mathrm{\infty }}{lim}|\frac{k}{k(1+\frac{1}{k})}|$
$=|\frac{x^2}{4}|\underset{k\to \mathrm{\infty }}{lim}|\frac{1}{1+\frac{1}{k}}|$
$=|\frac{x^2}{4}$