Comparison and Limit Comparison Tests — Question 7

Section 11-7 · Calculus pages 2073–2074

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*Question 7 Consider ∑n=1∞3n+15n−2\displaystyle\sum_{n=1}^{\infty}\frac{3^n+1}{5^n-2}.

  1. Select a geometric benchmark from the dominant exponential terms.

  2. Compute the Limit Comparison Test ratio.

  3. Classify the series and identify the benchmark’s common ratio.

*Question 7 – Solution

The dominant quotient is 3n/5n=(3/5)n3^n/5^n=(3/5)^n. Let bn=(3/5)nb_n=(3/5)^n, whose common ratio 3/53/5 has absolute value less than 11; hence ∑bn\sum b_n converges.

(3n+1)/(5n−2)(3/5)n=1+3−n1−2⋅5−n→1.\frac{(3^n+1)/(5^n-2)}{(3/5)^n} =\frac{1+3^{-n}}{1-2\cdot5^{-n}}\longrightarrow1. The original terms and benchmark are positive for n≥1n\ge1, and the ratio limit is finite and positive.

The two series have the same convergence behavior. Since the benchmark is geometric and convergent, ∑n=1∞3n+15n−2 converges.\boxed{\sum_{n=1}^{\infty}\frac{3^n+1}{5^n-2}\text{ converges}.}