MY NOTES $\newline$ ASK YOUR TEACHER $\newline$ PRACTICE ANOTHER $\newline$ 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.) $\newline$ $\sum_{n=0}^{\infty}(2 n) \cdot\left(\frac{x}{8}\right)^{n}$

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Q. MY NOTES

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ASK YOUR TEACHER

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PRACTICE ANOTHER

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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}(2 n) \cdot\left(\frac{x}{8}\right)^{n}

Ratio Test Application: To find the interval of convergence for the power series $\sum_{n=0}^{\infty}(2n)\left(\frac{x}{8}\right)^{n}$ , we will use the Ratio Test, which states that for a series $\sum a_n$ , if the limit $L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$ exists, then the series converges if $L < 1$ , diverges if $L > 1$ , and is inconclusive if $L = 1$ .
Identifying Terms: First, we identify the terms $a_n = (2n)\left(\frac{x}{8}\right)^{n}$ and $a_{n+1} = (2(n+1))\left(\frac{x}{8}\right)^{n+1}$ . We will plug these into the Ratio Test formula.
Calculating Limit: Now, we calculate the limit $L = \lim_{n \to \infty} \left|\frac{(2(n+1))\left(\frac{x}{8}\right)^{n+1}}{(2n)\left(\frac{x}{8}\right)^{n}}\right|$ .
Simplifying Expression: Simplify the expression inside the limit: $L = \lim_{n \to \infty} \left|\frac{(2(n+1))\left(\frac{x}{8}\right)^{n+1}}{(2n)\left(\frac{x}{8}\right)^{n}}\right| = \lim_{n \to \infty} \left|\frac{2(n+1)}{2n} \cdot \frac{x}{8}\right|$ .
Limit as n Approaches Infinity: Further simplification gives us: $L = \lim_{n \to \infty} \left|\frac{n+1}{n} \cdot \frac{x}{8}\right| = \lim_{n \to \infty} \left|\left(1 + \frac{1}{n}\right) \cdot \frac{x}{8}\right|$ .
Inequality for Convergence: As $n$ approaches infinity, $\frac{1}{n}$ approaches $0$ , so the limit simplifies to $L = \left|\frac{x}{8}\right|$ .
Checking Endpoints: For the series to converge, we need $L < 1$ , which means $\left|\frac{x}{8}\right| < 1$ . Solving this inequality gives us $-8 < x < 8$ .
Substitute x = $-8$ : Now we need to check the endpoints $x = -8$ and $x = 8$ to see if the series converges at these points. We do this by substituting $x = -8$ and $x = 8$ into the original series and checking for convergence.
Substitute x = $8$ : Substitute $x = -8$ into the original series: $\sum_{n=0}^{\infty}(2n)\left(\frac{-8}{8}\right)^{n}$ . This simplifies to $\sum_{n=0}^{\infty}(2n)(-1)^{n}$ , which is a divergent series because the terms do not approach zero as $n$ approaches infinity.
Final Interval of Convergence: Substitute $x = 8$ into the original series: $\sum_{n=0}^{\infty}(2n)\left(\frac{8}{8}\right)^{n}$ . This simplifies to $\sum_{n=0}^{\infty}(2n)$ , which is also a divergent series because the terms increase without bound as $n$ approaches infinity.
Final Interval of Convergence: Substitute $x = 8$ into the original series: $\sum_{n=0}^{\infty}(2n)\left(\frac{8}{8}\right)^{n}$ . This simplifies to $\sum_{n=0}^{\infty}(2n)$ , which is also a divergent series because the terms increase without bound as $n$ approaches infinity.Since the series diverges at both endpoints, the interval of convergence is $(-8, 8)$ , using interval notation.