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Series

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Zahidul Islam

1. The document discusses sequences and series in mathematics. 2. It provides examples of sequences, such as the sequence 1, -1, 1, -1, ..., and discusses whether sequences converge or diverge. 3. It also examines various infinite series, testing whether they converge or diverge using tools like the Ratio Test.

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e.9., t, 5,7, ..., t' ,Ir, A sequence is a nth term in. a sequence 'For examPle, if un By A ls a limit I then rrye r{rite sequence, e,g,; 7; I : l, -1, A sbquence which '. 3. l; -t, t, -1, ;., "
H"t: H L: t: , r, l l' I t, : i,r ., t. ;r' l, : 1'.t20 Which bf.rhe lnllnite 20.9 n+ PRO Ans. Convergeiit :, ; ':l ' il:, Anrl Divergent ,l l- Th .( {i 2. If REMEMBER THE20.6 (vD 0) : (i, (rv) =.Ql -Errl .Seh Sh.t I*q( &r .0 for;all val.ues of .r ,lgglnl}r/' = - ' limrx'=0if,r<ln+6 log a or' ./ , .i ,I r 'l .I , ,, J serii'i I , Solt ': ,. H( Er calledls a )
t: ,il ,,] l ri 'l ili "l. :i. rl !l i i -'ll:': ieries Hence, S, does not tend Solution. Here, .- .:. x rr. s,
L, .1'.|22 I r- Hence, (D) When r let r
1f Hence,,the given series is convergent when P > 1. Case2:P=f When p,= 1, the given seiies becomes 1 +-:. 1- -+ t6 On adding (l), (2), (3) and (4), we get given se-ries is L=k r1.P or Proof. By definition of both posltivo >m., ,tli : i l:
..{i$ 0n Case I rf.' r trar. .-k< 1' + !.
l. ${.j:,,; ' l:i., i. ,16 and non-zero, ;. Iun and Ivr, conveiie or .'. Iy, I diverge together ..l since f, v, , = )-a is of . ..,. .n2. 1 2 : tl*r;.'"".r 1128 lnrirtite Series ) ilo = +l+ l+ Let us compare I zn wilh I Y, , where -1- :':, i' un yn I l+ I ,.:,, I -n ++ 1 I"€t Iv" converg€ or ylknPthe form withp+2>1-
nfirfite ' ' (M.9:U:.2ooo) Here, we have ralt:i i . J .. r.'i,r.l .r:.:,1r. fj ",,. l'.. .: 2._+_+ l+2-t l+Z-2 5 ___+... l+2'' He;e ilr= Let us compare 'Examplc 12. Exanine the Solution. Her"e, we have r'ilr I"., = lim }'+4. uilx.iJl ,, Arr *.,r- ;l r-t/t*; n I ?r = _T: n2, I n lnfinite Series Here ,. 1179 Ans.' i form f, a ,r .f,.ii ,.'i (M.D.U.'2oAJ) Solution. ',. ; . !$.1 . 4o= l+ : ', '-,
EXERCISE 20.5 Examinc lhe converBence or divergencc of the follbwing series: , ,- , * 3a * 1.* * q.+ *.... Ans. convergent " -. 2'4 3 42 4 43 1.2 7.2.3 7.2.3.4' ...@ Ans. convergent2' 1+ 13 * r33 * l-:.s.2 lnfinite Series .1 D' $ 2n3 +,5 *r4" t iAne.Convergent t,i..: _ril il, ' ' . -,,r i. 1231 -+-+-+...@ -' 1.2 3.4 5.6 111/ L+-+-*.........@ '' r,2.3 2.3.4 3.4.5 1 131, Ans.,Diverginti' I 'r' Ans. Convergent .(M.D. IJniversity' Dec' 2004) t: '. . Ans, Convergent 'ti Ans, ConveJgent ., Ans. Convergent' .+ 22 ol,[. "afiE& ,5. 6. :, I = ,,(' + I Jr+{r+1 .an 3t + Divergent > a, convergenl; il x 3 a, DivergentAns. If r 9. Ans. Cunvergent R]TTIO TEST 6_', 12. Z r/{n' * t; - n ', Ans Divergent' . A!rs. lo"pyergent r ',,r::",, ' .. l:r r iaTl Ans. 2n + T;14. Ir.l 15. 16. L;n,.1r (l +r Strtcmcnl. Il2 ur ls a PositUe lerm thcnel (t) thc seiles ls 8. 11. 13. 10. itr.l 4'+no +J;L-n2 +l,.1 Casc By' I I I j i , i I j Convergent =k ,,:, { '.:: t .+..., @ ll2 t *-,'l . i
limiq n ofBy ) Coruider the t+o Thus, froqr (1) and (2) Hence, wheri Thus, ritiq
Whictr is finite and ,.i,. ' .,: series whose n'h term is ;2n we have un= -,n' BY D'Aiembertrs Ratio Tegt +l n9 2 rl lr Ur+l 4o I :,, t+ (4+ 1++

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Series

  • 1. e.9., t, 5,7, ..., t' ,Ir, A sequence is a nth term in. a sequence 'For examPle, if un By A ls a limit I then rrye r{rite sequence, e,g,; 7; I : l, -1, A sbquence which '. 3. l; -t, t, -1, ;., "
  • 2. H"t: H L: t: , r, l l' I t, : i,r ., t. ;r' l, : 1'.t20 Which bf.rhe lnllnite 20.9 n+ PRO Ans. Convergeiit :, ; ':l ' il:, Anrl Divergent ,l l- Th .( {i 2. If REMEMBER THE20.6 (vD 0) : (i, (rv) =.Ql -Errl .Seh Sh.t I*q( &r .0 for;all val.ues of .r ,lgglnl}r/' = - ' limrx'=0if,r<ln+6 log a or' ./ , .i ,I r 'l .I , ,, J serii'i I , Solt ': ,. H( Er calledls a )
  • 3. t: ,il ,,] l ri 'l ili "l. :i. rl !l i i -'ll:': ieries Hence, S, does not tend Solution. Here, .- .:. x rr. s,
  • 5. 1f Hence,,the given series is convergent when P > 1. Case2:P=f When p,= 1, the given seiies becomes 1 +-:. 1- -+ t6 On adding (l), (2), (3) and (4), we get given se-ries is L=k r1.P or Proof. By definition of both posltivo >m., ,tli : i l:
  • 7. l. ${.j:,,; ' l:i., i. ,16 and non-zero, ;. Iun and Ivr, conveiie or .'. Iy, I diverge together ..l since f, v, , = )-a is of . ..,. .n2. 1 2 : tl*r;.'"".r 1128 lnrirtite Series ) ilo = +l+ l+ Let us compare I zn wilh I Y, , where -1- :':, i' un yn I l+ I ,.:,, I -n ++ 1 I"€t Iv" converg€ or ylknPthe form withp+2>1-
  • 8. nfirfite ' ' (M.9:U:.2ooo) Here, we have ralt:i i . J .. r.'i,r.l .r:.:,1r. fj ",,. l'.. .: 2._+_+ l+2-t l+Z-2 5 ___+... l+2'' He;e ilr= Let us compare 'Examplc 12. Exanine the Solution. Her"e, we have r'ilr I"., = lim }'+4. uilx.iJl ,, Arr *.,r- ;l r-t/t*; n I ?r = _T: n2, I n lnfinite Series Here ,. 1179 Ans.' i form f, a ,r .f,.ii ,.'i (M.D.U.'2oAJ) Solution. ',. ; . !$.1 . 4o= l+ : ', '-,
  • 9. EXERCISE 20.5 Examinc lhe converBence or divergencc of the follbwing series: , ,- , * 3a * 1.* * q.+ *.... Ans. convergent " -. 2'4 3 42 4 43 1.2 7.2.3 7.2.3.4' ...@ Ans. convergent2' 1+ 13 * r33 * l-:.s.2 lnfinite Series .1 D' $ 2n3 +,5 *r4" t iAne.Convergent t,i..: _ril il, ' ' . -,,r i. 1231 -+-+-+...@ -' 1.2 3.4 5.6 111/ L+-+-*.........@ '' r,2.3 2.3.4 3.4.5 1 131, Ans.,Diverginti' I 'r' Ans. Convergent .(M.D. IJniversity' Dec' 2004) t: '. . Ans, Convergent 'ti Ans, ConveJgent ., Ans. Convergent' .+ 22 ol,[. "afiE& ,5. 6. :, I = ,,(' + I Jr+{r+1 .an 3t + Divergent > a, convergenl; il x 3 a, DivergentAns. If r 9. Ans. Cunvergent R]TTIO TEST 6_', 12. Z r/{n' * t; - n ', Ans Divergent' . A!rs. lo"pyergent r ',,r::",, ' .. l:r r iaTl Ans. 2n + T;14. Ir.l 15. 16. L;n,.1r (l +r Strtcmcnl. Il2 ur ls a PositUe lerm thcnel (t) thc seiles ls 8. 11. 13. 10. itr.l 4'+no +J;L-n2 +l,.1 Casc By' I I I j i , i I j Convergent =k ,,:, { '.:: t .+..., @ ll2 t *-,'l . i
  • 10. limiq n ofBy ) Coruider the t+o Thus, froqr (1) and (2) Hence, wheri Thus, ritiq
  • 11. Whictr is finite and ,.i,. ' .,: series whose n'h term is ;2n we have un= -,n' BY D'Aiembertrs Ratio Tegt +l n9 2 rl lr Ur+l 4o I :,, t+ (4+ 1++