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

k=1k(1+k2)p

Short Answer

Expert verified

Theintegralx=1x1+x2pdxconvergesforp>1.Thus,theseriesk=1k(1+k2)pisconvergentforp>1.

Step by step solution

01

Step 1. Given information is:

k=1k(1+k2)p

02

Step 2. Examining nature of given function:

Considerthefunctionf(x)=x1+x2p.Thefunctionf(x)=x1+x2piscontinuous,decreasing,withpositiveterms.Thereforealltheconditionsofintegraltestarefulfilled.So,integraltestisapplicable.

03

Step 3. Solving the integral:

Considertheintegral:x=1f(x)dx=x=1x1+x2pdx.Therefore,x=1f(x)dx=limkx=1kx1+x2pdx=12limku=2k2+1duupPut1+x2=u,2xdx=du=12limku-p+1-p+12k2+1(Integrating)=12(-p+1)limk1up-12k2+1(Substitution)

04

Step 4. Result:

Theimproperintegralconvergestofinitevalueonlywhenp>1.Therefore,theintegralx=1x1+x2pdxconvergesforp>1.Thus,theseriesk=1k(1+k2)pisconvergentforp>1.

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