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Tentukan jari-jari konvergen deret berikut dengan cara formula Cauchy Hadamard: (bobot 20) sum_(n=2)^(oo)(3^(n))/((n-2))(z)^(n)

55. Tentukan jari-jari konvergen deret berikut dengan cara formula Cauchy Hadamard: (bobot 2020)\newline∑n=2∞3n(n−2)(z)n \sum_{n=2}^{\infty} \frac{3^{n}}{(n-2)}(z)^{n}

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Q. 55. Tentukan jari-jari konvergen deret berikut dengan cara formula Cauchy Hadamard: (bobot 2020)\newline∑n=2∞3n(n−2)(z)n \sum_{n=2}^{\infty} \frac{3^{n}}{(n-2)}(z)^{n}
  1. Use Cauchy-Hadamard formula: Use the Cauchy-Hadamard formula to find the radius of convergence, RR. The formula is 1R=lim⁡n→∞∣an∣1n\frac{1}{R} = \lim_{n \to \infty} |a_n|^{\frac{1}{n}}, where ana_n is the nth term of the series.
  2. Identify nth term: Identify the nth term of the series, an=3n(n−2)a_n = \frac{3^n}{(n-2)}.
  3. Apply formula: Apply the Cauchy-Hadamard formula: 1R=lim⁡n→∞∣3n(n−2)∣1n\frac{1}{R} = \lim_{n \to \infty} \left|\frac{3^n}{(n-2)}\right|^{\frac{1}{n}}.
  4. Simplify expression: Simplify the expression: 1R=lim⁡n→∞(3nn(n−2)1n)\frac{1}{R} = \lim_{n \to \infty} \left(\frac{3^{\frac{n}{n}}}{(n-2)^{\frac{1}{n}}}\right).
  5. Calculate radius: Since 3(n/n)3^{(n/n)} is just 33 and (n−2)(1/n)(n-2)^{(1/n)} approaches 11 as nn approaches infinity, we have 1R=31\frac{1}{R} = \frac{3}{1}.
  6. Calculate radius: Since 3(n/n)3^{(n/n)} is just 33 and (n−2)(1/n)(n-2)^{(1/n)} approaches 11 as nn approaches infinity, we have 1/R=3/11/R = 3/1. Therefore, R=1/3R = 1/3.

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