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In Exercises 48–51 find all values of p so that the series converges.

k=1lnkkp

Short Answer

Expert verified

Theintegralx=1lnxxpdxconvergesforp>1.Thus,theseriesk=1lnkkpisconvergentforp>1.

Step by step solution

01

Step 1. Given information is:

k=1lnkkp

02

Step 2. Examining nature of given function:

Considerthefunctionf(x)=lnxxp.Thefunctionf(x)=lnxxpiscontinuous,decreasing,withpositiveterms.Thereforealltheconditionsofintegraltestarefulfilled.So,integraltestisapplicable.

03

Step 3. Solving the integral:

Considertheintegral:x=1f(x)dx=x=1lnxxpdx.Therefore,x=1f(x)dx=limkx=1klnxxpdx=limku=0lnkueu1-pduPutlnx=u,1xdx=du

04

Step 4. Result:

Theimproperintegralconvergestofinitevalueonlywhenp>1.Therefore,theintegralx=1lnxxpdxconvergesforp>1.Thus,theseriesk=1lnkkpisconvergentforp>1.

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