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Does ∑k=1[infinity]​(k+2)!(−1)k+1(k3k)​ converge absolutely, conditionally or not at all?

User Slf
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1 Answer

1 vote

Final answer:

The series
∑k=1[infinity]​(k+2)!(−1)k+1(k3k)​

Step-by-step explanation:

The series
∑k=1∞(k+2)!(−1)k+1(k3k)

To determine this, we can use the ratio test. Taking the absolute value of the terms of the series, we have
∑k=1∞(k+2)!(−1)k+1(k3k)

Using the ratio test, we have
(k+3)!k3(k+3)/((k+2)!k3k) = (k+3)k3equal to ∞, which means the series diverges. However, when we alternate the signs of the terms, the series converges conditionally.

User Jamuhl
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