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Let 0<p<1. Show that 0≤1klnk≤1kpfor large values of k. Explain why we cannot use a p-series with0<p<1in a comparison test to verify the divergence of the series∑k=2∞1klnk.

Short Answer

Expert verified

The comparison test provides no information on the series∑k=2∞1k(lnk) divergence.

Step by step solution

01

Step 1. Given information

An expression is given as0≤1klnk≤1kp

02

Step 2. Verification

For larger values of k ,

kp≤kkPlnk≤klnk1klnk≤1kplnk1kplnk<1kp1klnk<1kp

For 0<p<1, the p-series is divergent. If the series ∑k=1xbkconverges, then the series ∑i=1∞akconverges, according to the comparison test.

But the series ∑i=1∞bk=∑k=1∞1kpis divergent, so we cannot find the value of convergence.

The comparison test provides no information on the series' divergence.

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