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Determine whether the series is absolutely convergent, conditionally convergent, or divergent:

ₙ₌₁∑[infinity](-1)ⁿ -1/n+2

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Final answer:

The series is conditionally convergent because it passes the Alternating Series Test. The series is not absolutely convergent, as the absolute value of its terms forms a divergent p-series with p=1.

Step-by-step explanation:

A student asked to determine whether the series ℓ₁∞∑((-1)^ℓ -1)/(ℓ+2) is absolutely convergent, conditionally convergent, or divergent.

To determine the type of convergence, we begin with the Alternating Series Test (AST). If the series terms, b_n = 1/(ℓ+2), are positive, decreasing, and approaching zero as ℓ approaches infinity, the series passes the AST and is at least conditionally convergent.

The terms 1/(ℓ+2) are indeed positive and decreasing. Additionally, as ℓ approaches infinity, 1/(ℓ+2) approaches zero. Therefore, the series passes the AST and is conditionally convergent.

To investigate absolute convergence, consider the absolute value of the series terms, |((-1)^ℓ -1)/(ℓ+2)|, which simplifies to 1/(ℓ+2). We must then use a comparison test or ratio test to determine whether this new series converges. Since 1/(ℓ+2) is a p-series with p=1, which is known to diverge, the absolute value of the original series does not converge, indicating the series is not absolutely convergent but is conditionally convergent.

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