find the radius of convergence, r, of the series.\n∑(n = 2 to ∞) (x^n + 6)/(7n!)\nfind the interval, i, of…

Explanation:

Step1: Apply Ratio - Test formula

The Ratio - Test states that for a series $\sum_{n = 2}^{\infty}a_{n}$, we consider $\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_{n}}\right|$, where $a_{n}=\frac{x^{n}+6}{7n!}$ and $a_{n+1}=\frac{x^{n + 1}+6}{7(n + 1)!}$. So, $\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_{n}}\right|=\lim_{n\rightarrow\infty}\left|\frac{\frac{x^{n+1}+6}{7(n + 1)!}}{\frac{x^{n}+6}{7n!}}\right|$.

Step2: Simplify the ratio

We know that $(n + 1)!=(n + 1)\times n!$. So, $\lim_{n\rightarrow\infty}\left|\frac{\frac{x^{n+1}+6}{7(n + 1)!}}{\frac{x^{n}+6}{7n!}}\right|=\lim_{n\rightarrow\infty}\left|\frac{x^{n+1}+6}{7(n + 1)!}\cdot\frac{7n!}{x^{n}+6}\right|=\lim_{n\rightarrow\infty}\left|\frac{x^{n+1}+6}{x^{n}+6}\cdot\frac{1}{n + 1}\right|$. Since $\lim_{n\rightarrow\infty}\frac{1}{n + 1}=0$ and $\left|\frac{x^{n+1}+6}{x^{n}+6}\right|$ is bounded for all real - valued $x$, we have $\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_{n}}\right| = 0$ for all $x\in R$.

Step3: Determine the radius of convergence

By the Ratio - Test, if $\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_{n}}\right|=L$, the radius of convergence $R$ is given by $R=\frac{1}{L}$ when $0\lt L\lt\infty$, $R = 0$ when $L=\infty$, and $R=\infty$ when $L = 0$. Since $L = 0$, $R=\infty$.

Step4: Determine the interval of convergence

Since the radius of convergence $R=\infty$, the interval of convergence $I$ is $(-\infty,\infty)$.

Answer:

$R=\infty$ $I=(-\infty,\infty)$