11–27. Alternating Series Test Determine whether the following series converge. ∑ (k = 1 to ∞) (−1)ᵏ⁺¹ k¹/ᵏ
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Identify the general term of the series: \(a_k = (-1)^{k+1} k^{1/k}\). This is an alternating series because of the factor \((-1)^{k+1}\), which causes the terms to alternate in sign.
Recall the Alternating Series Test (Leibniz Test), which states that an alternating series \(\sum (-1)^k b_k\) converges if two conditions are met: (1) the sequence \(b_k\) is positive, decreasing, and (2) \(\lim_{k \to \infty} b_k = 0\).
Set \(b_k = k^{1/k}\) (the absolute value of the terms without the alternating sign). Check if \(b_k\) is positive and decreasing for sufficiently large \(k\). Since \(k^{1/k} > 0\) for all \(k\), the positivity condition is satisfied.
Analyze the limit \(\lim_{k \to \infty} k^{1/k}\). Use the fact that \(k^{1/k} = e^{(\ln k)/k}\). Evaluate the limit of the exponent \(\lim_{k \to \infty} \frac{\ln k}{k}\), which approaches 0, so the limit of \(k^{1/k}\) approaches \(e^0 = 1\).
Since \(\lim_{k \to \infty} b_k = 1 \neq 0\), the Alternating Series Test fails, and therefore the series does not converge by this test.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Alternating Series Test
The Alternating Series Test determines the convergence of series whose terms alternate in sign. It requires that the absolute value of the terms decreases monotonically to zero. If these conditions hold, the series converges, even if it does not converge absolutely.
For any infinite series to converge, the limit of its general term as k approaches infinity must be zero. If the terms do not approach zero, the series diverges. This is a necessary condition for convergence but not sufficient on its own.
The term k^(1/k) represents the k-th root of k, which approaches 1 as k becomes very large. Understanding this limit helps determine whether the terms of the series decrease to zero, a key step in applying the Alternating Series Test.