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Use the Ratio Test to determine whether the series is convergent or divergent. sum_(n=1)^(oo)(-1)^(n-1)(3^(n))/(2^(n)n^(3)) Identify a_(n)

Use the Ratio Test to determine whether the series is convergent or divergent.\newline∑n=1∞(−1)n−13n2nn3 \sum_{n=1}^{\infty}(-1)^{n-1} \frac{3^{n}}{2^{n} n^{3}} \newlineIdentify an a_{n}

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Full solution

Q. Use the Ratio Test to determine whether the series is convergent or divergent.\newline∑n=1∞(−1)n−13n2nn3 \sum_{n=1}^{\infty}(-1)^{n-1} \frac{3^{n}}{2^{n} n^{3}} \newlineIdentify an a_{n}
  1. Identify general term: Identify ana_{n} as the general term of the series, which is an=(−1)n−1(3n)/(2nn3)a_{n} = (-1)^{n-1}(3^{n})/(2^{n}n^{3}).
  2. Apply Ratio Test: Apply the Ratio Test by finding the limit of the absolute value of an+1an\frac{a_{n+1}}{a_n} as nn approaches infinity.
  3. Calculate an+1a_{n+1}: Calculate an+1a_{n+1} which is (−1)n(3n+1)/(2n+1(n+1)3)(-1)^n(3^{n+1})/(2^{n+1}(n+1)^3).
  4. Find absolute value: Now, find the absolute value of an+1/an=∣(−1)n(3n+1)2n+1(n+1)3⋅2nn3(−1)n−1(3n)∣a_{n+1}/a_{n} = \left|\frac{(-1)^{n}(3^{n+1})}{2^{n+1}(n+1)^{3}} \cdot \frac{2^{n}n^{3}}{(-1)^{n-1}(3^{n})}\right|.
  5. Simplify expression: Simplify the expression to get \left|\frac{\(3\)}{\(2\)}\left(\frac{n^{\(3\)}}{(n+\(1\))^{\(3\)}}\right)\right|.
  6. Take limit: Take the limit as \(n approaches infinity of ∣32(n3(n+1)3)∣\left|\frac{3}{2}\left(\frac{n^{3}}{(n+1)^{3}}\right)\right|.
  7. Calculate final limit: Since the degrees of the polynomials in the numerator and denominator are the same, the limit is the ratio of the leading coefficients. So, the limit is 32\frac{3}{2}.

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