Lecture Note
University
California State UniversityCourse
MATH 150B | Calculus IIPages
5
Academic year
2023
yung dump
Views
0
10.2 Infinite Series 53 * An infinite series is the sum of an in finite sequence of a, + a2 + a3 + + an + ... = numbers n=) where. an is the nth term of the series S1 = a, is the 1st partial sum of the series S2 = a, + a2 is the 2 nd partial sum of the series n th Sn = a, + az two + an = ak is the n partial sum of K=1 series. If the sequence of partial sums converges to a limit L then we say the series converges: and we write a, + az + ... + an +1.. = an = L If the sequence of partial sums of n=1 the Serie does not converge, then we say the series diverges. E XP I + 1/2 2 + 1/4 + 1/8 + 1/6 + (t)" Partial sums n=1 I 1/2 4 $ First S1=1 = 2-1 second 2 S2 = I + 1/2 =2- 1 Third S3 = It 1/2 + 1/4 = 2- 1/4 n-1 : nth Sn = 1+ t/2 + 1/1 + + (1) = 2- Note that this sequence of partial sums converges to 2 because lim Sn = 2 - O = & . n-s . Thus, we say the 00 sum of the in finite series ( (1) = = = n = 1
Geometric Series 54 Geometric series are series of the form: 8 n-1 n-1 a + ar + ar2 + wit ar + = * where a and are fixed real numbers n= a to r is called the ratio and can be positive : n-1 1+ 1/2 + 1/1 two + (1)) v=1/2 / a = 1 or negative : n-1 To determine the convergence 1-1 - + 1 9 and divergence (-1) of + the geometric V=-L 3 , 9=1 series / we consid 12 If r=1 / then the nth partial sum of the 3 cases: n-1 geometric series is Sn = a + a (1) + a(1)? + a(1)=na and the series diverges because lim 00 Sn = + 8 depending on the sign of a. 2. If r=-1, then the nth partial sum of the series is Sn = of - a + - al + a + ac-15- n-1 = so if n even a if n odd Thus, the series diverges because the nth partial sum alternate between a and O. 3 If V# I and rf -) (i.e |r|#1) then we can determine the convergence or divergence as follows! Sn = a + ar + ar 2 + X + ar n-2 n-1 rsn = Lv + If ar + av + + n-1 + ar n + ar Sn - rsn = a - ar" (=> Sn (1-r) = a (1-r") Sn = a (1-r") / v#1 1-r
If 1rl < 1, then r" o as n x 55 w n-1 lim Sn = a Thus ar converges to n->00 1-r n=1 a If Irl >1 then r" as and the series diverges If/KI, the geometric series at ar + n-1 converges to a : n-1 a 1-r Zar n=1 to If 1/1/1,1, the series diverges Exp It n-1 00 1/2 + 2 1/4 + n-1 + \ = =2 n=1 n-1 00 00 Exp 1-1-1-14 +(1/) Earth n-1 = E(-1) I 3 = n=1 n=1 Exp: Express the reputing decimal numbers as the ratio of two integers: 23 23 23 0.23 = 0.23 23 23 ... + I 100 (100) (100) 23 100 (too) 23 23 n=1 Too = = 23 = 1/00 99 99 100 100 2 0.7 = 0.777 7 7 = + 10 (10)2 7 7 7 = 10 = 1- 1/9 3 0.06 =0.0666 6 6 = + 100 1000 10000 6 6 100 = = 15 = 100 = 90 9 1- to
The nth Term Test for Divergent Series 56 * If lim an fails to exist or is different from n->00 D zero, then E an diverges N=1 & Exp 1 The series E n+1 diverges because lim an = lim n+1 = 1 n n=1 n-soo nas to 2 The sentes EN in diverges because lim an = lim nn = as n=1 now n-sas 3 The series (-1)"+ diverges because lim (-1) n+1 does notexist no n=1 4 The series as -n diverges because lim -n = - 1/2 # 6 2n+5 n->00 2n+5 NII Ih 7 If Ean an converges, then lim an = O . n w n=1 Note that Th7 does not say that if 0 DO lim an =0 then an converges : exp E 1/4 "harmonic" series n->so n=\ n=1 w Th 8 If " an = A and E bn = B are convergent series; then ASI 11 Sum Rule : E (ant bn) = Ean + E bn = A+B 2 Difference Rule E (an - bn) = Ean - Ebn = A- B 3 Constant Multiple Rule, E K an = K Ean = KA, KEIR Note that I Every nonzevo constant multiple of divergent series is Diverges [2] If Ean converges and Ebn diverges, then E (antbn) and ((an-bn) - both diverges.
"Telescoping Series " 57 Exp Find a formula for the nth partial sum of the following series and use it to determine if the series converges or diverges. If the series converges find the sum. 8 1 E 1/1/2 - /14 n=1 Sn = (1 tr) V2 3 + ( (yu) +115 + (You in) \ ( K - tun) Sn = 1- Vn+1 => n->00 lim Sn = 1. . Thus, the series w converges to 1 i.e n=1 E (1/- / = 1 2 n=1 n(n+1) " we use partial fraction" n n(n+1) = As A + B A=1 it , n-l -17 = ) + (1-1) the (*) 5/4/4 B=-1 Sn = 1- 1/1 lim n->00 Sn = 1. . Thus, the series converges to 1. i.e ninth \ = 1 n(n+1) H=1 Exp Find the sum of the fo llowing series & E/N + 1/1) = (5+1) t ( /2/ + 1/ 1+(++1)+ 1/2 two n=o = [5+ + 5/5 4 + upla +1.1 + [It 1/3 + 1/ + +1/+++++] = 5 I + 1-1 1-1/3 3 = 5 + 1/2 3 = 10 + 2/2 = 23 2
Infinite Series
Please or to post comments