No SIM 11:48 AM 86% bartleby.com = bartleby Q&A Math / Calculus / Calculu.../8th Edit... Chapter 11, Problem 19RE VUIU ulatIVII, 1 3-5.. (2n - 1) Σ The given series c аn — 5"n! n=1 n=1 1-3.5. (2n-1) where a n = 5" n! Then, the (n+ 1) th term is, 1-3.5..(2(n+1)-1) an1= S"+1 (n+1)! 1-3-5..(2n-1)2n+1) an+1 5"+1 (n+1) Obtain the limit of an+1 аn 1-3.5..(2n-1)(2n+1) 5n+1 (nt1)! аn+1 lim = lim 1-3.5..(2n-1) аn n oo п—0 5П n! 1-3.5...(2n-1)2n+1) 5" n! = lim 5"+n+1)! 1-3-5...(2n-1) n oo (2n+1)n! = lim 5(п+1)n! п— 00 n(2+) = lim 5n(1+ n oo Simplify further and apply the limit, (2+) аn+1 lim = lim 5(1+) аn n oo n oo 2+
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For this problem how did they went from:
1.3.5...(2n+1) to 1.3.5....(2n-1)(2n+1) for finding an+1 ?
Step by step
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