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91Ó°ÊÓ

Use the ratio test for absolute convergence to determine whether the series in Exercises 30–35 converge absolutely or diverge.
∑k=0∞(-1)k1.3.5...(2k+1)1.4.7...((3k+1)

Short Answer

Expert verified

The series ∑k=0∞(-1)k1.3.5...(2k+1)1.4.7...((3k+1)converges absolutely.

Step by step solution

01

Step 1. Given Information.

The series:

∑k=0∞(-1)k1.3.5...(2k+1)1.4.7...((3k+1)

02

Step 2. Rewrite the series.

ak=(-1)k1.3.5...(2k+1)1.4.7...((3k+1)

03

Step 3. Find ak+1.

ak+1=(-1)k+11.3.5...(2(k+1)+1)1.4.7...(3(k+1)+1)=(-1)k+11.3.5...(2k+3)1.4.7...((3k+4)

04

Step 4. Calculate ak+1ak.

ak+1ak=(-1)k1.3.5...(2k+3)1.4.7...((3k+4)(-1)k1.3.5...(2k+1)1.4.7...((3k+1)=(-1)2k+33k+4=2k+33k+4

05

Step 5. Take limits.

limk→∞ak+1ak=limk→∞((2k+3)3k+4)=limk→∞(k(2+3k)k(3+4k))=(23)=23<1

So by the ratio test, the series converges absolutely.

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