Determining Convergence or Divergence
In Exercises 17–46, use any method to determine whether the series converges or diverges. Give reasons for your answer.
∑(from n=1 to ∞) [(2 + (−1)ⁿ) / 1.25ⁿ]

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Identify the general term of the series: $a_n = \frac{2 + (-1)^n}{1.25^n}$.

Recognize that the denominator \$1.25^n$ is an exponential term with base greater than 1, which tends to grow as $n$ increases.

Consider the behavior of the numerator $2 + (-1)^n$, which oscillates between $2 + 1 = 3$ (for even $n$) and $2 - 1 = 1$ (for odd $n$), so it remains bounded.

Since the numerator is bounded and the denominator grows exponentially, compare the series to a geometric series with ratio $r = \frac{1}{1.25} < 1$.

Conclude that because the terms behave like a geometric series with ratio less than 1, the series converges by the Comparison Test or by recognizing it as a sum of geometric series.

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주요 개념

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Geometric Series and Its Convergence

A geometric series has the form ∑ arⁿ, where r is the common ratio. It converges if |r| < 1 and diverges otherwise. Understanding this helps analyze series with terms involving exponential expressions like 1.25ⁿ.

Alternating Series and the Alternating Series Test

An alternating series has terms that alternate in sign, often involving (−1)ⁿ. The Alternating Series Test states such a series converges if the absolute value of terms decreases monotonically to zero. This is useful when the series includes (−1)ⁿ factors.

Comparison and Limit Comparison Tests

These tests compare a given series to a known benchmark series to determine convergence or divergence. The Limit Comparison Test uses the limit of the ratio of terms to decide if both series behave similarly, aiding in analyzing complex series.

Determining Convergence or DivergenceIn Exercises 17–46, use any method to determine whether the series converges or diverges. Give reasons for your answer.∑(from n=1 to ∞) [(2 + (−1)ⁿ) / 1.25ⁿ]