we will use the ratio test to determine the radius of convergence. we have lim(n -> ∞) |(a_n + 1)/(a_n)| =…

Explanation:

Step1: Simplify the ratio of terms

Simplify $\lim_{n\rightarrow\infty}\left|\frac{\frac{(x - 11)^{n+1}}{(n + 1)^{2}+1}}{\frac{(x - 11)^{n}}{n^{2}+1}}\right|$. First, use the rule of dividing fractions $\frac{a/b}{c/d}=\frac{a\cdot d}{b\cdot c}$. So we get $\lim_{n\rightarrow\infty}\left|\frac{(x - 11)^{n+1}(n^{2}+1)}{(x - 11)^{n}((n + 1)^{2}+1)}\right|$. Then, by the property of exponents $\frac{a^{m}}{a^{n}}=a^{m - n}$, $(x - 11)^{n+1}/(x - 11)^{n}=x - 11$. And $(n + 1)^{2}+1=n^{2}+2n+1 + 1=n^{2}+2n+2$. So the limit becomes $\lim_{n\rightarrow\infty}\left|(x - 11)\frac{n^{2}+1}{n^{2}+2n+2}\right|$.

Step2: Find the limit as $n\rightarrow\infty$

Divide both the numerator and denominator of $\frac{n^{2}+1}{n^{2}+2n+2}$ by $n^{2}$: $\frac{n^{2}+1}{n^{2}+2n+2}=\frac{1+\frac{1}{n^{2}}}{1+\frac{2}{n}+\frac{2}{n^{2}}}$. As $n\rightarrow\infty$, $\frac{1}{n}\rightarrow0$ and $\frac{1}{n^{2}}\rightarrow0$. So $\lim_{n\rightarrow\infty}\frac{1+\frac{1}{n^{2}}}{1+\frac{2}{n}+\frac{2}{n^{2}}}=1$. Then $\lim_{n\rightarrow\infty}\left|(x - 11)\frac{n^{2}+1}{n^{2}+2n+2}\right|=\left|x - 11\right|$.

Step3: Determine the radius of convergence

For the series to converge by the Ratio - Test, $\left|x - 11\right|\lt1$. The general form for the radius of convergence $R$ from $\left|x - a\right|\lt R$ (where $a = 11$ in our case) gives $R = 1$.

Answer:

$1$