HW13 - Math 191 11.3 2 converges a geometric series with r = 1 e HW 13 Solutions <1 11.3 6 converges 1-2 =-2 which is a convergent p-series(p = 3 2 3\/2

# HW13 - Math 191 11.3 2 converges a geometric series with r...

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Math 191HW 13 Solutions11.3: 2 converges; a geometric series withr=1e<111.3: 6 converges;Xn=1-2nn=-2Xn=11n3/2, which is a convergent p-series (p=32).11.3: 18 diverges;limn→∞an= limn→∞1 +1nn=e6= 011.3: 28 diverges by the Integral Test;Z1xx2+ 1dx;u=x2+ 1du=2xdx12Z2du4=limb→∞12lnub2=limb→∞12(lnb-ln 2) =11.3: 39a.Z2dxx(lnx)p;u=lnxdu=dxxZln 2u-pdu= limb→∞u-p+1-p+ 1bln 2= limb→∞(11-p)[b-p+1-(ln 2)-p+1]=1p-1(ln 2)-p+1, p >1, p <1the improper integral converges ifp >1 and diverges ifp <1.Forp= 1;Z2dxxlnx= limb→∞[ln(lnx)]b2= limb→∞[ln(lnb)-ln(ln 2)] =, so the improper integraldiverges ifp= 1.b. Since the series and the integral converges or diverges together,Xn=21n(lnn)pconverges if and onlyifp >1.11.4: 4 converges by the Direct Comparison Test;1+cosnn22n2and the p-series Σ1n2converges.