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

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

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 $f(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2^{n}(n+1)}(x-1)^n$.

Step 3 ：The radius of convergence $R$ is given by the reciprocal of the limit of the absolute ratio of consecutive terms, i.e., $R = \frac{1}{\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|}$, 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 $\lim_{n\to\infty} \left|\frac{-1}{2} \cdot \frac{n+1}{n+2}\right| = \frac{1}{2}$.

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 $f(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2^{n}(n+1)}(x-1)^n$ and its radius of convergence $R = 2$.