Then rn is rational for each n, but lim rn = √ 2 is irrational Problem 4 (7 5) (c) lim (√4n2 + n − 2n) Solution (a) √ n2 + 1 − n = 1 √ n2 +1+ n → 0 (b) √ sn+1 = lim n→∞ √ sn +1= √ s + 1, and this equation has solution s = (1 + √
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[PDF] HW Solutions (Math 323)
8 6) a) If sn → 0, then for ∈ > 0 we have sn −0 = sn < ∈ for all n>N Consider sn−0 = sn 2n − √4n2 + n 4(√4n2 + n + 2n)∣∣∣∣∣ = √ 1+1/4n − 1 4(√1+1/4n + 1) Thus, sn + tn > M for all n>N, showing that lim(sn + tn) = ∞
[PDF] Math 2260 Exam Practice Problem Solutions
4 (3 4 )n−1 = 3/4 1 - 3/4 = 3/4 1/4 = 3 Therefore, ∞ ∑ n=1 2n + 3n 4n ( −1)n √ 1+n n2 (x - 2)n ∣ ∣ ∣ = lim n→∞ / 2 + n x - 2n+1 (n + 1)2 Since (2n + 2)(2n +1)=4n2 + 6n + 2 and since (n + 1)2 = n2 + 2n + 1, the above limit
[PDF] MATH 4310 :: Introduction to Real Analysis I :: Spring 2015 :: Langou
n2 + 1 − n is of the form (∞−∞) which is not determined and makes the problem 1 n + 2) = 4, so lim( 1 √ 4 + 1 n + 2 ) = 1 4 , lim( √ 4n2 + n − 2n) = 1 4
[PDF] HOMEWORK 3 - UCLA Math
Then rn is rational for each n, but lim rn = √ 2 is irrational Problem 4 (7 5) (c) lim (√4n2 + n − 2n) Solution (a) √ n2 + 1 − n = 1 √ n2 +1+ n → 0 (b) √ sn+1 = lim n→∞ √ sn +1= √ s + 1, and this equation has solution s = (1 + √
[PDF] Math 104: Introduction to Analysis SOLUTIONS Alexander Givental
8 8 c Prove that lim[√4n2 + n − 2n]=1/4 Put sn = an/n and find that sn+1/sn = a/(n + 1) tends to 0 as n → ∞ Therefore, by the previous exercise, limsn = 0
[PDF] SOLUTIONS OF HOMEWORK-3 (1) (a) 1 4n2 − 1 = 1 2 [ 1 2n − 1
2n+1 ] Hence sn → 1 2 So, ∑ n 1 4n2−1 = 1 2 (b) The partial sum sn = 2[1 + 1 But for x=-1, the series converges to −∞ Since for x < 1 both the series ∑ n 2n ≤ 2 (n+1)2 , so the series converges (c) 1 + √ n (1 + n)3 − 1 ≤ 1 + nn Now compute Limn→∞ an+1 an = Limn→∞2( n n+1 )n+1 = Limn→∞
[PDF] EXERCISES 2 – SEQUENCES - Euler
(1 + (-1)n+1) ] Limits of Sequences 5 Find the limit lim n→∞ 4n3 + 5n2 - 7n + 2 (2n - 1)((2n + 1)2n - 2n) = 4n2 + 2n + 1 4n2 = 1 9 Find the limit lim n→∞ (-2 )n+3 e−4 14 Find the limit lim n→∞ (2n + 3 2n - 1 )3n+2 Solution The given n→∞ √ n + √ n + / n / n + 1 [1] l) lim n→∞ 7ncos2 n (2n + 1)(2n - 1) [0]
[PDF] MATH 314 Assignment 1 (a) Prove that limn→∞ √ n = ∞ and
n6 Consequently, limn→∞ bn = −∞ The sequence diverges (c) We have cn = 3n + 2n 3n − 4n = 3n(1 + 2n 3n ) −4n(1 − 3n 4n ) = − (3 4 )n 1+(2 3 )n
[PDF] Mathematical analysis Sequences IV—Limits October 2013
√ n3−1)cos n n2+6 ; 74 limn→∞(1 + 2 n )
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