Find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval. If the interval of convergence is an interval, enter your answer using interval notation. If the interval of convergence is a finite set, enter your answer usina set notation.) $\newline$ $\sum_{n=0}^{\infty}(2 n) !\left(\frac{x}{8}\right)^{n}$

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Q. Find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval. If the interval of convergence is an interval, enter your answer using interval notation. If the interval of convergence is a finite set, enter your answer usina set notation.)

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\sum_{n=0}^{\infty}(2 n) !\left(\frac{x}{8}\right)^{n}

Use Ratio Test: To find the interval of convergence for the power series $\sum_{n=0}^{\infty}\frac{(2n)!}{8^n}x^n$ , we will use the ratio test, which states that a series $\sum a_n$ converges if the limit $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$ is less than $1$ .
Find General Term: First, we find the ratio $\left| \frac{a_{n+1}}{a_n} \right|$ for our series. The general term of our series is $a_n = \frac{(2n)!}{8^n}x^n$ .
Calculate Ratio: Now, we find $a_{n+1}$ , which is $\frac{(2(n+1))!}{8^{n+1}}x^{n+1}$ .
Limit Calculation: We calculate the ratio $\left| \frac{a_{n+1}}{a_n} \right|$ as follows: $\newline$ $\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{(2(n+1))!}{8^{n+1}}x^{n+1}}{\frac{(2n)!}{8^n}x^n} \right| = \left| \frac{(2(n+1))!}{(2n)!} \cdot \frac{8^n}{8^{n+1}} \cdot \frac{x^{n+1}}{x^n} \right| = \left| \frac{(2n+2)(2n+1)}{8} \cdot x \right|.$
Check Endpoints: Taking the limit as $n$ approaches infinity, we get: $\newline$ $\lim_{n \to \infty} \left| \frac{(2n+2)(2n+1)}{8} \cdot x \right| = \lim_{n \to \infty} \left| \frac{2n(2n+1)}{8} \cdot x \right| = \lim_{n \to \infty} \left| \frac{4n^2 + 2n}{8} \cdot x \right| = \lim_{n \to \infty} \left| \frac{n^2(4 + \frac{2}{n})}{8} \cdot x \right|.$
Evaluate at x= $0$ : Since $n$ is approaching infinity, the term $\frac{2}{n}$ becomes negligible, and we can simplify the limit to: $\newline$ $\lim_{n \to \infty} \left| \frac{n^2(4)}{8} \cdot x \right| = \lim_{n \to \infty} \left| \frac{n^2}{2} \cdot x \right| = \infty.$ $\newline$ This means that the ratio test gives us an inconclusive result, as the limit does not exist.
Determine Interval: Since the ratio test was inconclusive, we need to check for convergence at the endpoints of the interval. The endpoints are determined by the radius of convergence, which in this case is $0$ , since the limit as $n$ approaches infinity is not finite. Therefore, the endpoints to check are $x = 0$ .
Determine Interval: Since the ratio test was inconclusive, we need to check for convergence at the endpoints of the interval. The endpoints are determined by the radius of convergence, which in this case is $0$ , since the limit as $n$ approaches infinity is not finite. Therefore, the endpoints to check are $x = 0$ .At $x = 0$ , the series becomes $\sum_{n=0}^{\infty}\frac{(2n)!}{8^n}0^n$ , which is simply $0$ for all terms except for $n = 0$ . This means the series converges at $x = 0$ .
Determine Interval: Since the ratio test was inconclusive, we need to check for convergence at the endpoints of the interval. The endpoints are determined by the radius of convergence, which in this case is $0$ , since the limit as $n$ approaches infinity is not finite. Therefore, the endpoints to check are $x = 0$ .At $x = 0$ , the series becomes $\sum_{n=0}^{\infty}\frac{(2n)!}{8^n}0^n$ , which is simply $0$ for all terms except for $n = 0$ . This means the series converges at $x = 0$ .Since the series only converges at $x = 0$ , the interval of convergence is a single point, which we write in set notation as $\{0\}$ .