Sf1.pdf - Calculus Math 21C Spring 2018 Sample Final...

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Calculus : Math 21C, Spring 2018 Sample Final Questions: Solutions 1.Do the following sequences{an}converge or diverge asn→ ∞?If asequence converges, find its limit. Justify your answers. (a)an=2n2+ 3n3(b)an=sin(n2)n;(c)an=nlnn. Solution. 1
2.Do the following series converge or diverge? State clearly which test youuse.(a)Xn=1n+ 46n-17(b)Xn=1rnn4+ 7(c)Xn=1(-5)n+1(2n)! (d)(-1)nlnnn(e)114+124-134+144+154-164+174-194+· · ·(f)Xn=1en-en+1 Solution. We write each series as c n . 2
(c) We have c n +1 c n = ( - 5) n +2 / (2 n + 2)! ( - 5) n +1 / (2 n )! = 5(2 n )! (2 n + 2)! = 5 (2 n + 1)(2 n + 2) . It follows that lim n →∞ c n +1 c n = 0 . Since this limit exists and is less than 1, the series converges absolutely by the ratio test, and therefore it converges. (d) For x 3, we have ln x > 1, and d dx ln x x = 1 - ln x x 2 < 0 , so ln n/n is decreasing for n 3, and ln n/n 0 as n → ∞ by l’Hˆ ospital’s rule. Hence, the series converges by the alternating series test. (e) We have | c n | = 1 n 4 so the series converges absolutely, since the p -series with p = 4 con- verges. Therefore the series converges since any absolutely convergent series is convergent. (f) Either note that c n = e n - e n +1 = e n (1 - e ) diverges to -∞ as n → ∞ , so the series diverges by the n th term test. Or note that the series is a telescoping series of the form c n = b n - b n +1 with b n = e n and the limit of b n as n → ∞ does not exist, so the series is a divergent telescoping series. 3

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