find all the values of x such that the given series would converge.\n\n sum _ { n = 1 } ^ { infty } \frac {…

Explanation:

Step1: Apply the ratio test

Let (a_{n}=\frac{2^{n}(x - 2)^{n}(n + 1)}{n+8}). Then (a_{n+1}=\frac{2^{n+1}(x - 2)^{n+1}(n + 2)}{n+9}). The ratio is (\left|\frac{a_{n+1}}{a_{n}}\right|=\left|\frac{2^{n+1}(x - 2)^{n+1}(n + 2)}{n+9}\cdot\frac{n + 8}{2^{n}(x - 2)^{n}(n + 1)}\right|). Simplify: (\left|\frac{a_{n+1}}{a_{n}}\right|=\left|2(x - 2)\cdot\frac{n+2}{n + 1}\cdot\frac{n + 8}{n+9}\right|). As (n\to\infty), (\lim_{n\to\infty}\frac{n+2}{n + 1}=1) and (\lim_{n\to\infty}\frac{n + 8}{n+9}=1). So (\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\left|2(x - 2)\right|).

Step2: Find the convergence condition

For convergence, (\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|\lt1). (\left|2(x - 2)\right|\lt1). Divide both sides by (2): (\left|x - 2\right|\lt\frac{1}{2}). This gives (-\frac{1}{2}\lt x - 2\lt\frac{1}{2}). Add (2) to all parts: (2-\frac{1}{2}\lt x\lt2+\frac{1}{2}), i.e., (\frac{3}{2}\lt x\lt\frac{5}{2}).

Step3: Check the endpoints

• When (x=\frac{3}{2}): The series becomes (\sum_{n = 1}^{\infty}\frac{2^{n}(\frac{3}{2}-2)^{n}(n + 1)}{n+8}=\sum_{n = 1}^{\infty}\frac{2^{n}(-\frac{1}{2})^{n}(n + 1)}{n+8}=\sum_{n = 1}^{\infty}\frac{(- 1)^{n}(n + 1)}{n+8}). (\lim_{n\to\infty}\frac{n + 1}{n+8}=1\neq0). By the divergence test, the series diverges at (x=\frac{3}{2}).
• When (x=\frac{5}{2}): The series becomes (\sum_{n = 1}^{\infty}\frac{2^{n}(\frac{5}{2}-2)^{n}(n + 1)}{n+8}=\sum_{n = 1}^{\infty}\frac{2^{n}(\frac{1}{2})^{n}(n + 1)}{n+8}=\sum_{n = 1}^{\infty}\frac{n + 1}{n+8}). (\lim_{n\to\infty}\frac{n + 1}{n+8}=1\neq0). By the divergence test, the series diverges at (x=\frac{5}{2}).

Answer:

(\left(\frac{3}{2},\frac{5}{2}\right))