Question

2 nº-1 Find the sum of the series [m=1 in=1 (nº +1)

Answer 1

(n2-12 (n27) 2 na Uni n2 2 E Un= n?(1-12) =) Un = (12-1) m24)? na (1+ynz)? (1-1/m2) Cityne) • Take Ilna Ynz-, then lim Mori ntolles a -Hence by comparision test both times E1%8 (12402 Convergent → Sum of the sends is finite (n2-12 The C22412² z (121 are ich le <o: CS Scanned with CamScanner

Now we find sum of the series

(m - 1) (m + 1) (+1) 2 m=1 1 ਨੂੰ ( ) m2 +1 (2 +12 Split the series: 1 (2 +1)2 1 ਇੰਜ ) - ( ਆ) + m2 +1 =1 m=1 2 (2 +1)2

Split the series: 1 η2 +1 η=1 2 (η2 + 1)2 O 1 1 +Σ η2 +1 m2 +1 n=0 n=0 2 + Σ (m2 + 1)2 η=1 Since the bounds are finite, the number of terms is finite as well, and we just calculate the sum by summing up the terms.

1 + η2 +1 =0 ΣΕΙΣ η2 +1 n=0 n=1 2 (-1) + (m2 + 1)2 8 1 + Σ 1 η2 +1 n=0. η=1 2 (η2 +1)? is a known series. η2 +1 n=0 It is 1 a2 +12 n=0 1 1 coth (πα) + 02 with a 1. Therefore,

+ w 2+2+2 2n + 2i 2n – 2i n=1 1 1 + + 2(n + i)? 2(n − 1)2 Split the series: 1 1 + coth (7) + s. i + 2n + 2i 2n – 2i n=1

1 + ਵਿਚ ' )) 2(n + i)? 2(n − 1) 1 TT + ä сoth (7) 2 1 + (Σ + n=1 2(n+i)? 1 8 + n=1 2(n − 1)2 i (2+2 2n + 2i 2n – 21 n=1 Pull the constant out of the series: 1 2 coth () NI 1 + 3+) + n=1 2(n+3)

1 + n=1 2(n − 1)2 i 2 Σ + 2n + 2i 2n 2i n=1 1 IT = coth (a (a + 2 2 1 + + 2 (n+i)? n=1 8. 1 + n=1 2(n − 1)2 . (+) 2n + 2i 2n – 2i n=1 Split the series: 1 1 Nie + 2 2 2 n=1 (n+i)? TT 1 + -coth (T) + 2 n=1 2(n − 1)? +

i t ( 2+2 2n + 2i 2n - 2i n=1 2 0 1 1 1 + E into + Σ 2 n=0 i)? n=0 (nti (n + coth () + coth TT 1 + 2 n=1 2(n − 1)2 + Ź Ź + WI 2n + 2i 2n 21 n=1 Since the bounds are finite, the number of terms is finite as well, and we just calculate the sum by summing up the terms.

1 1 1 + Σ 2 (η + 1)? n=0 1 1 π + + coth 2 2 n=0 (n + i)? 1 (π) +Σ + n=1 2(η – )? Σ( 4 * ) 2n + 20 η=1 1 + (1) 2 + 2 2 1 π + 2 ο coth (π) + n=0 (n + 1)? 1 Σ + n=1 2(η – 1)? ο Θ. Σ + 2n + 20 2η 21 n=1 1 is a known series. n=0 (η + 1)?

It is Σ(n + 1) " = ζ (n, r), n=0 η>1 with η = 2, η = 1. Therefore, 1 1 (Σ) - ξενα 2 2 1 + n=1 2(η – 1)? n=1 (π) + Σ Στο ) Scoth (π) + Σ Σ( (ζ (2, i)) 2 + 1 + η=1 2(η - - i)? , ο + 2n + 2 2η - 21 n=1

Pull the constant out of the series: 5 (2, i) IT + -coth (a 2 2 1 + (Σ + n=1 2(n – i) i (122 2n + 2i 2n 2i n=1 TT 5 (2, i) 2 + coth (1) 2 1 + + 2 (n – i) n=1 i i ( 12+2 2n + 2i 2n 2i n=1 Split the series: 1 1 Ś (2, i) (.) 4420 2 n=1 (n − 1)2 TT + coth (1) + 2

(2n + 3 + 2x - 3) n=1 0 1 1 1 = + Σ 2 n=0 (m - 1)? (n − 3)? n=0 T + Ś (2, 1) 2 +coth (T) + 2 i (+3+2) 2n + 2i n=1 Since the bounds are finite, the number of terms is finite as well, and we just calculate the sum by summing up the terms. 1 2 n=0 1 Σ + 2 (n − 1)2 5 (2, 2) + 2 n= (n − 1)? +=coth (71) + 2 TT i + 2n + 2i 2n – 21 n=1

1 1 (1) 2 + 2 n=0 (m - 1)? + 513,0 coth () + ( 7+23+27 - ) i n=1 1 is a known series. n=0 (n – i)? It is (n +r)-" = 5 (n,r), n=0 n > 1 with n = 2, r = -1. Therefore, 1 1 + Σ 2 2 n=0 (n (n – i) TT + Ś (2, 1) 2 + ż co coth (a) + & (+*+2) 2n + 2i n=1

1 = + 2 2 Ś (2,1) TT + + coth (71) + 2 2 i ( ****2-2) 2n + 2i n=1 Can't find the exact value. The series is convergent. 1 Ś (2, i) 1 + +55 (2,-2) 2 2 TT + coth (TT 2 + ( 22 ) 2n + 2i - 2i n=1 ~(-1.07168656798994 + 3.0 1 + -coth (a 2 2 5 (2, i) +55 (2, -2) 2 2 1 +

Thus, na – 1 n=1 (n2 + 1)2 -0.571686567989943 Ś (2, 1) + -coth () + 2 2 TT + 3.0 - 10-24; +36(2, -i -i) Answer: na – 1 n=i (n2 + 1) -0.571686567989943 TT 5 (2, ) + coth (n) + 2 2 + 3.0 1 10-21; + (2, -i) 2 ~ 0.467987576101402