Finding the radius R of convergence for a power series, Single Variable Calculus
We do four (not three) examples to find the radius of convergence R for different power series using the ratio test. (Previous video: • Power Series Interval of Convergence,... )
Example 1: Power Series ∑𝑥^𝑛/(4𝑛−1). Applying the ratio test, we get L=|𝑥|. Therefore, the series converges for |𝑥| less than 1 and diverges for |𝑥| great than 1. We manually check 𝑥=−1 and 𝑥=1. The series converges for 𝑥=−1 but diverges for 𝑥=1, leading to an interval of convergence [−1,1) and a radius of convergence 𝑅=1.
Example 2: Power Series with natural log ∑ (−1)^𝑛𝑥^𝑛/(3^𝑛 ln(𝑛)). Applying the ratio test, we obtain: L=|𝑥|/3. The series converges for |𝑥| less than 3 and diverges for |𝑥| greater than 3. Testing 𝑥=−3 and 𝑥=3, we find that it diverges for 𝑥=−3 but converges for 𝑥=3. Thus, the interval of convergence is (−3,3] with a radius of convergence 𝑅=3.
Example 3: Power Series with factorial ∑ 𝑛!(5𝑥−1)^𝑛. First, we rewrite this power series as ∑𝑛!5^𝑛(𝑥−1/5)^𝑛, identifying the coefficients as 𝑛!5^𝑛 and the center as 1/5. Using the ratio test, we find 𝐿=∞ (for 𝑥≠1/5). The series only converges at its center, 𝑥=1/5. Thus, the radius of convergence is 0, and the interval of convergence is the single point {1/5}.
Example 4: Power Series ∑𝑥^𝑛/𝑛^𝑛. Applying the ratio test, 𝐿=0 (for all 𝑥). This series converges for all 𝑥, implying an infinite radius of convergence. Therefore, the interval of convergence is (−∞,∞).
Revisit the Ratio Test here: • Absolute Convergence and the Ratio Te...
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