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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 thatRn10-6

k=11k2

Short Answer

Expert verified

Part (a):

Thevalueoftheintegralisx=11x2dx=1Theintegralconverges.Thus,theseriesk=11k2isconvergent.

Part (b):

S10=11+122+132+...+1102=1+0.25+0.11+0.0625+...+0.011.5487Therefore,the10thterminthesequenceofpartialsumstoapproximatethesumoftheseriesisS101.5487

Part (c):

localid="1649210871544" ThetenthremainderR10,isboundedby:0R10x=10x-2dx.....(1)x=10x-2dx=0.1,therefore,0R10110Therefore,theboundon10thremainder,R10is0.1

Part (d):

S10LS10+B101.5487L1.5487+0.1(Substitution)1.5487L1.6487Therefore,theintervalcontainingthesumoftheseriesisL(1.5487,1.6487).

Part (e):

na(x)dx10-61n10-6n1000000Therefore,thevalueofnsothatRn10-6holdsisn=1000000

Step by step solution

01

Step 1. Given information is:

k=11k2

02

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

Considerthefunctionf(x)=1x2.Thefunctionf(x)=1x2iscontinuous,decreasing,withpositiveterms.Thereforealltheconditionsofintegraltestarefulfilled.So,integraltestisapplicable.

03

Part (a) Step 2. Solving the integral: 

Considertheintegral:x=1f(x)dx=x=11x2dx.Therefore,x=1f(x)dx=limkx=1k1x2dx=limk-1x1k(Integrating)=limk-1k+1(Substitution)=0+1=1

04

Part (a) Step 3. Result

Thevalueoftheintegralisx=11x2dx=1Theintegraldiverges.Thus,theseriesk=11k2isconvergent.

05

Part (b) Step 1. Finding S10

ThevalueofS10is:S10=11+122+132+...+1102=1+0.25+0.11+0.0625+...+0.011.5487Therefore,the10thterminthesequenceofpartialsumstoapproximatethesumoftheseriesisS101.5487

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=10x-2dx.....(1)Thevalueoftheintegralx=10e-xdxis:x=10x-2dx=0.1(Valueofintegralsolvedabovein(a))Therefore,equation(1)becomes:0R101100R100.1(Simplify)Therefore,theboundon10thremainder,R10is0.1

07

Part (d) Step 1. Finding Interval

AssumeB100.1andk=1a(k)=L.UsingtheresultSnk=1a(k)Sn+Bn,thefollowinginequalityisobtained.S10LS10+B101.5487L1.5487+0.1(Substitution)1.5487L1.6487Therefore,theintervalcontainingthesumoftheseriesisL(1.5487,1.6487).

08

Part (e) Step 1. Finding n

Tofindthesmallestvalueofn,findnsuchthatfollowingholds:na(x)dx10-61n10-6n1000000Therefore,thevalueofnsothatRn10-6holdsisn=1000000

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