Question

Let 5,=5- 7n be the nth partial sum of Σας Select all that apply 131 n = 1 Your answer: lim a,= 0 os,5 n = 1 91(n-1)-7n 13" (s,-5) is convergent. n = 1 7n 0 an=5 –. 13" Σ È (,+())+5+ n=1 Lan=5 n=1 snis divergent by the Divergence Test. n=1

Answer 1

Given, 7n s = 5- be the nth partial sum of Σα, . n 13" n=1 Then, a = , +a, +az + ax + az +... n=1 So, Si = a S2 = a + az Sz = a + a2 + az Sn-1 = a + a2 + az + ... +0,-1 Sn = a + a, +az +...+an Then,

-5+ 13n-1 an = Sn -Sn-1 7n 7(n-1) = 5- 13" 7(n-1) in 13-1 13" 13x7(n-1) - 7n 13" 91(n-1) – 7n 13" = Now find the limit of a,. liman = = lim n00 n00 = lim 91(n-1)-7n 13" 91n-91-7n 13" 84n-7 13" = lim n 88 form = lim 84 13" log (13) 100 = = 0 lim an = 0 n00

7n ŠS - EAST 13" N=1 n=1 Apply divergent test, 7n lim 5 = 570 13" Divergent 7n (s» –5)= 1 5 5 13" N=1 7n Σ 13" n=1 By ratio test, the serics & 1 is convergent. → Ž(s, –5)is convergent. n=1

Σα (5)]-Σ 91(n-1) - 7η 13" + (5) 5 -Σ (91(n-1) - 7η 13" + 1=1 Yi=1 5 =0 + 2 5 – 2 Hence, 5. ΙΣ α, + 5 5 7 2 Y=1