Chuỗi đan dấu và tiêu chuẩn Leibniz?

Oct 2, 2014 ... Let. an=(5n−3n37n3+2)n . By Root Test,. limn→∞n√|an|=limn→∞n√∣∣∣(5n−3n37n3+2)n∣∣∣. by cancelling the nth-root and the nth-power,. (B) lim=<1. 11-10 e. (C) lim-. +1. <1. })-00 e e. (D) lim. <1 e. <1. (E) lim (n+1)! ... Y+3= -1(x-1). Y = -1(x-1)-3 at x=1.5 y = = 1/2-3 7/½. ( 312 - 312) m = 3-3/2 ... ( ) = lim n n k=1 t2. 0 k( 3) = 1-lim n. 1n( 3). Thus we obtain the lemma. D. For simplicity, we put. 1n := 0 1n := 1n( 3) ( = 1 2 ). Recall n( ) = n( ; 3 2) ... Feb 4, 2021 ... \. \. \. \. 2n + 1. 5n + 4. −. 2. 5. \. \. \. \. = \. \. \. \. 10n + 2 − (10n + 8). 5(5n + 4. \. \. \. \. = 6. 25n + 20. <. 6. 25n. < . So lim ... The series is alternating. II. |a. |. |a | for n. 1. III. lim. →. Question #5 Evaluating ∑∞ n=1 22n−1 5n+1 Answer: You can rewrite it ... 3 √2 1 2 1 2√3 1 2 √2 2 . Since ℎ′ (x) − = 1 1 2√x−2 , then ℎ′ (3) ... with r= 2/3 < 1, so converges. original series converges. 3) & n²-5n n=6 n² ... lim n²-sn t = lim 1-5/n 1. 2-Sn converges (p-series p=2), the original. (b) an = = n! lim. 818. (-1)". 2" (n + 2). To earn the third point a response must correctly use limit notation to evaluate the improper integral, find an evaluation of 0 e (or 1), and conclude that ... Sep 19, 2015 ... 2 Answers 2 ... And limn→∞(12)1n=limn→∞31n=1. So the answer is 23. However I assume you are a beginner like me, thus maybe prefer to prove it by ...