Determine end behavior of polynomial and rational functions

Full solution

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 using set notation.)

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

Ratio Test: To find the interval of convergence of the power series $\sum_{n=0}^{\infty} \frac{(2n)!}{8^n} x^n$ , we will use the ratio test, which states that for a series $\sum a_n$ , if $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L$ , then the series converges if $L < 1$ , diverges if $L > 1$ , and is inconclusive if $L = 1$ .
Calculate Expressions: First, we need to find the expression for $a_n$ and $a_{n+1}$ . In our case, $a_n = \frac{(2n)!}{8^n} x^n$ and $a_{n+1} = \frac{(2(n+1))!}{8^{n+1}} x^{n+1}$ .
Limit Calculation: Now, we calculate the limit of the ratio $\left| \frac{a_{n+1}}{a_n} \right|$ as $n$ approaches infinity: $\newline$ $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(2(n+1))!}{8^{n+1}} x^{n+1} \cdot \frac{8^n}{(2n)!} x^{-n} \right|$
Simplify Expression: Simplify the expression inside the limit: $\newline$ $\lim_{n \to \infty} \left| \frac{(2(n+1))!}{8^{n+1}} x^{n+1} \cdot \frac{8^n}{(2n)!} x^{-n} \right| = \lim_{n \to \infty} \left| \frac{(2n+2)!}{(2n)!} \cdot \frac{1}{8} \cdot x \right|$
Factorial Simplification: Further simplify the factorial expression: $\newline$ $\lim_{n \to \infty} \left| \frac{(2n+2)!}{(2n)!} \cdot \frac{1}{8} \cdot x \right| = \lim_{n \to \infty} \left| (2n+2)(2n+1) \cdot \frac{1}{8} \cdot x \right|$
Limit Analysis: Since both $2n+2$ and $2n+1$ grow without bound as $n$ approaches infinity, the limit of their product will also approach infinity. Therefore, the limit of the ratio test is infinite for any non-zero value of $x$ : $\newline$ $\lim_{n \to \infty} \left| (2n+2)(2n+1) \cdot \frac{1}{8} \cdot x \right| = \infty$ $\newline$ This means that the series only converges when $x = 0$ .
Limit Analysis: Since both $2n+2$ and $2n+1$ grow without bound as $n$ approaches infinity, the limit of their product will also approach infinity. Therefore, the limit of the ratio test is infinite for any non-zero value of $x$ : $\newline$ $\lim_{n \to \infty} \left| (2n+2)(2n+1) \cdot \frac{1}{8} \cdot x \right| = \infty$ $\newline$ This means that the series only converges when $x = 0$ .Since the series only converges for $x = 0$ , the interval of convergence is a single point, which we express using set notation: $\newline$ $\text{Interval of convergence} = \{0\}$