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Chapter 7: Q. 20 (page 652)

Give an example of convergent series ∑k=1∞akand ∑k=1∞bksuch that the series ∑k=1∞akbkdiverges.

Short Answer

Expert verified

∑k=1∞akbk=∑k=1∞1kdiverges

Step by step solution

01

Step 1. Given information

∑k=1∞akand∑k=1∞bkconverges

02

Step 2. Check if ∑k=1∞akand ∑k=1∞bk converges

∑k=1∞ (−1)k+1akand ∑k=1∞ (−1)kakboth converge

Therefore,

∑k=1∞ ak=(−1)kkconverges

Now,

∑k=1∞ bk=(−1)kkdiverges

03

Step 3. Check if ∑k=1∞ak bk diverges

∑k=1∞ akbk=∑k=1∞ (−1)kk⋅(−1)kk=∑k=1∞ 1k

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Prove Theorem 7.31. That is, show that if a function a is continuous, positive, and decreasing, and if the improper integral ∫1∞a(x)dxconverges, then the nth remainder, Rn, for the series∑k=1∞a(k) is bounded by0≤Rn=∑k=n+1∞a(k)≤∫n∞a(x)dx

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