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Section 12.3 nz, . 7 - 20. f(z) = —z is continuous and positive for a > 2, and f’(x) = cea <0 for x > 2, so f is decreasing . zr “—" : “Ing Inz 1)° H — I = ing _ing 1 H inn ln I = dz jm z 1] : [by parts] = 1. Thus, ) a = converges by the Integral Test. . n=1 n=? 30. (a) f(z) = 1/2‘ is positive and continuous and f(z) = —4/z° is negative for 2 > 0, and so the Integral _ ee A 1 1 I 1 Test applies. 2nd & 610 = pate t ett ype © 1.082087. ~~ . i jf . 1 1 \_1 . = Rao < [ pide = jim [=a]. - a2 (-s5 + say) = Spgig o> the enna is at most 0.0003, sdeees Zz wi <8% = i not f Sao ss
so + saps S95 80+ BGs 1.082037 + 0.000250 = 1.082287 < s < 1.082037 + 0.000333 = 1.082370, so we get s + 1.08233 with error < 0.00005. (Ras [” de= ss So Rn < 0.00001 => oct = asi) 5 jf, ce 3n8° , 3n3 ~ 105 n> 3/(0)5/3 © 32.2, thatis, for n > 32. Section 12.4 . a a eae Sak for all n > 2, so Sa 1 Vii diverges by comparison with the divergent (partial) harmonic series = 1 22, Use the Limit Comparison Test with a, = aoe and bn = =: a, _ me(n+2 . 1+2 lm — = lim ( a lim o_ =1> 0. Since 4 —; is a convergent (partial) p-series mite by te (nts am (44) nas ™ ee nt2 =2>1),th te (p = 2 > 1), the series > ces also converges. Section 12.5 1 . i. 10. Yoon = D- -y Pe = xe 1)"bn. Now lim nb = him Sag = 5 HO: Since lim, an #0 n=1 (in fact the limit does not exist), the series diverges by the Test for Divergence. 
