Question

9–36. Comparison tests Use the Comparison Test or the Limit Comparison Test to determine whether the following series converge.∑ (k = 1 to ∞) 4ᵏ / (5ᵏ − 3)

Accepted Answer

Identify the given series: $$ \sum_{k=1}^{\infty} \frac{4^{k}}{5^{k} - 3} . We $$want to determine if this series converges using the Comparison Test or the Limit Comparison Test. Consider the behavior of the terms for large $$ k . $$Since $$ 5^{k} $$ grows much faster than 3, the denominator $$ 5^{k} - 3 $$ behaves approximately like $$ 5^{k} $$ for large $$ k . So $$the terms behave like $$ \frac{4^{k}}{5^{k}} . $$Simplify the approximate terms: $$ \frac{4^{k}}{5^{k}} = \left( \frac{4}{5} ight)^{k} . $$This is a geometric series with ratio $$ r = \frac{4}{5} $$, which is less than 1, so the geometric series $$ \sum \left( \frac{4}{5} ight)^{k} $$ converges. Use the Limit Comparison Test by computing $$ \lim_{k 	o \infty} \frac{\frac{4^{k}}{5^{k} - 3}}{\left( \frac{4}{5} ight)^{k}} . $$Simplify this limit to check if it is a finite positive number. If the limit is finite and positive, then by the Limit Comparison Test, the original series converges or diverges together with the geometric series $$ \sum \left( \frac{4}{5} ight)^{k} . $$Since the geometric series converges, conclude that the original series converges.

9–36. Comparison tests Use the Comparison Test or the Limit Comparison Test to determine whether the following series converge.

∑ (k = 1 to ∞) 4ᵏ / (5ᵏ − 3)

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Key Concepts

Comparison Test

Limit Comparison Test

Geometric Series and Dominant Term Analysis