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For each series in Exercises 44–47, do each of the following:

(a) Use the integral test to show that the series converges.

(b) Use the 10th term in the sequence of partial sums to approximate the sum of the series.

(c) Use Theorem 7.31 to find a bound on the tenth remainder R10.

(d) Use your answers from parts (b) and (c) to find an interval containing the sum of the series.

(e) Find the smallest value of n so that.

k=21k(lnk)2

Short Answer

Expert verified

Part (a):

Thevalueoftheintegralisx=21xlnx2=1ln2Theintegralconverges.Thus,theseriesk=21klnk2isconvergent.

Part (b):

S10=12(ln2)2+13(ln3)2+14(ln4)2+...+111(ln11)2=1.0406+0.2762+0.13+0.0772+...+0.01581.7002Therefore,the10thterminthesequenceofpartialsumstoapproximatethesumoftheseriesisS101.7002

Part (c):

ThetenthremainderR10,isboundedby:0R10x=101x(lnx)2dx.....(1)x=101x(lnx)2dx0.43429Therefore:0R100.43429Therefore,theboundon10thremainder,R10is0.43429

Part (d):

S10LS10+B101.7002L1.7002+0.43429(Substitution)1.7002L2.13449Therefore,theintervalcontainingthesumoftheseriesisL(1.7002,2.13449).

Part (e):

na(x)dx10-61lnn10-6lnn1000000Therefore,thevalueofnsothatRn10-6holdsisn=3x10434294

Step by step solution

01

Step 1. Given information is:

k=21k(lnk)2

02

Part (a) Step 1. Examining nature of given function: 

Considerthefunctionf(x)=1x(lnx)2.Thefunctionf(x)=1x(lnx)2iscontinuous,decreasing,withpositiveterms.Thereforealltheconditionsofintegraltestarefulfilled.So,integraltestisapplicable.

03

Part (a) Step 2. Solving the integral: 

Considertheintegral:x=2f(x)dx=x=21x(lnx)2dx.Therefore,x=21xlnx2dx=limkx=2k1xlnx2dx=limku=ln2lnk1u2duPutlnx=u,1xdx=du=limk-1uln2lnk=limk-1lnk+1ln2(Substitution)=0+1ln2=1ln2

04

Part (a) Step 3. Result

Thevalueoftheintegralisx=21xlnx2=1ln2Theintegralconverges.Thus,theseriesk=21klnk2isconvergent.

05

Part (b) Step 1. Finding S10

ThevalueofS10is:S10=12(ln2)2+13(ln3)2+14(ln4)2+...+111(ln11)2=1.0406+0.2762+0.13+0.0772+...+0.01581.7002Therefore,the10thterminthesequenceofpartialsumstoapproximatethesumoftheseriesisS101.7002

06

Part (c) Step 1. Finding bound on R10

Ifafunctionaiscontinuous,positive,anddecreasing,andiftheimproperIntegral1a(x)converges,thenthenthremainder,Rn,fortheseriesk=1a(k)isboundedby:0Rnna(x)dxFurthermore,ifBn=na(x)dx,thenSnk=1a(k)Sn+BnTheresultgivesthatthetenthremainderR10,isboundedby:0R10x=101x(lnx)2dx.....(1)Thevalueoftheintegralx=101x(lnx)2dxis:x=101x(lnx)2dx0.43429(Valueofintegralsolvedabovein(a))Therefore,equation(1)becomes:0R100.43429Therefore,theboundon10thremainder,R10is0.43429

07

Part (d) Step 1. Finding Interval

AssumeB100.43429andk=1a(k)=L.UsingtheresultSnk=1a(k)Sn+Bn,thefollowinginequalityisobtained.S10LS10+B101.7002L1.7002+0.43429(Substitution)1.7002L2.13449Therefore,theintervalcontainingthesumoftheseriesisL(1.7002,2.13449).

08

Part (e) Step 1. Finding n

Tofindthesmallestvalueofn,findnsuchthatfollowingholds:na(x)dx10-61lnn10-6lnn1000000Therefore,thevalueofnsothatRn10-6holdsisn=3x10434294

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