Find the Taylor series for $f$ centered at 1 and its radius of convergence $R$ if $f^{(n)}(1)=\frac{(-1{)}^{n}n!}{2^n(n+1)}$
$f(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{2^n(n+1)}(x-1{)}^n,R=\frac{1}{2}$
$f(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{2^n(n+1)}(x-1{)}^n,R=2$
None of the above
$f(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{2^n(n+1)}(x-1{)}^n,R=\mathrm{\infty }$
$f(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{2^n}{x}^n,R=2$

Step 1 ：The Taylor series for a function $f$ centered at $a$ is given by $f(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{f^{(n)}(a)}{n!}(x-a{)}^n$.

Step 2 ：Given that $f^{(n)}(1)=\frac{(-1{)}^{n}n!}{2^n(n+1)}$, we can substitute this into the Taylor series formula to get $f(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{2^n(n+1)}(x-1{)}^n$.

Step 3 ：The radius of convergence $R$ is given by the reciprocal of the limit of the absolute ratio of consecutive terms, i.e., $R=\frac{1}{\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|}$, where $a_n$ is the $n$th term of the series.

Step 4 ：In this case, $a_n=\frac{(-1{)}^n}{2^n(n+1)}$ and $a_{n+1}=\frac{(-1{)}^{n+1}}{2^{n+1}(n+2)}$.

Step 5 ：So, $\frac{a_{n+1}}{a_n}=\frac{(-1{)}^{n+1}/[2^{n+1}(n+2)]}{(-1{)}^n/[2^n(n+1)]}=\frac{-1}{2}\cdot \frac{n+1}{n+2}$.

Step 6 ：Taking the limit as $n$ approaches infinity, we get $\underset{n\to \mathrm{\infty }}{lim}|\frac{-1}{2}\cdot \frac{n+1}{n+2}|=\frac{1}{2}$.

Step 7 ：Therefore, the radius of convergence $R$ is the reciprocal of this limit, which is $R=\frac{1}{\frac{1}{2}}=2$.

Step 8 ：So, the Taylor series for $f$ centered at 1 is $f(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{2^n(n+1)}(x-1{)}^n$ and its radius of convergence $R=2$.