11.4 Overview

#6, p. 750

a[n_]:=(n-1)/(n^2*Sqrt[n])
b[n_]:=1/n^(3/2)
a[Range[10]]
b[Range[10]]
a[Range[10]]/b[Range[10]]
ListPlot[a[Range[10]]/b[Range[10]],Filling->Axis,PlotStyle->{Thick,PointSize[Large]}]

#16, p. 750

a[n_]:=1/(3n^4+1)^(1/3)
b[n_]:=1/n^(4/3)
a[Range[10]]
b[Range[10]]
N[a[Range[10]]/b[Range[10]]]
ListPlot[a[Range[10]]/b[Range[10]],Filling->Axis,PlotStyle->{Thick,PointSize[Large]}]

#28, p. 750

a[n_]:=Exp[1/n]/n
b[n_]:=1/n
a[Range[10]]
b[Range[10]]
a[Range[10]]/b[Range[10]]
ListPlot[a[Range[10]]/b[Range[10]],Filling->Axis,PlotStyle->{Thick,PointSize[Large]}]

#34, p. 750: one thing that we can say is that the error is less than the error made in the associated series b_n

a[n_]:=Sin[n]^2/n^3
a[Range[10]]
b[n_]:=1/n^3
Integrate[1/x^3,{x,n,Infinity}]
N[Total[a[Range[10]]]]
MaxError=N[1/(2*(10)^2)]
N[Total[a[Range[100]]]]
MaxError=N[1/(2*(100)^2)]
ListPlot[{a[Range[10]],b[Range[10]]},Filling->Axis,PlotStyle->{{Thick,PointSize[Large]},{Thick,PointSize[Large]}},PlotRange->{Automatic,{-.05,.15}}]