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n = 1 18 (−1)”+ ¹ 3 · 7 · 11 · · · (4n − 1)(x − 3)” · - 1)(x-3)" 4n

n = 1 18 (−1)”+ ¹ 3 · 7 · 11 · · · (4n − 1)(x − 3)” · - 1)(x-3)" 4n

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section: Chapter Questions
Problem 32T
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Find the interval of convergence of the power series. Also include a check for convergence at the endpoints of the interval.
. 33. (− 1)”+¹ 3 · 7 · 11 · · · (4n − 1)(x − 3)" • - - 4n 11=
Transcribed Image Text:. 33. (− 1)”+¹ 3 · 7 · 11 · · · (4n − 1)(x − 3)" • - - 4n 11=
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Follow-up Question

In this answer to the previous follow-up question, why do we multiply the (4n - 1) at the top? This was one of my previous follow-up questions, and the answer to that was "...no need to write (4n -1) in the top." And it is still confusing me.

Now, 을(-1)+1 3.7.11. - (4n-1)(x-3)^ n=1 կո an = 1-117+1 3.7. 11. կո = land = 3.7.11 կո (4n-1) (x-3)" (4n-1)/(x-3)^ &lantıl 3. 7. 11. - - 4n+1 Σ an (say) n=1 (4n-1) (4n+3) (x-3)/2+1 Series, Eland converges if M=1 By Ratio test lim lantil lanl < 1 L 1. at lim lantil 1, test fails. n-p 1 anl Now, lim lantil nto lanl = lim ntoo 37.11. 3.7.11. - =) lim lantil Flim (4n+3)1x-317+1 ntoo lanl 7-7700 4n+1 = lim lantıl n700 Ian' = lim (4n+3) 1x-31 n200 4 (4n-1) (4n+3) 1x-317+1 4n+1 (49-1) 1x-3/m կո x կո 1x-31"
Transcribed Image Text:Now, 을(-1)+1 3.7.11. - (4n-1)(x-3)^ n=1 կո an = 1-117+1 3.7. 11. կո = land = 3.7.11 կո (4n-1) (x-3)" (4n-1)/(x-3)^ &lantıl 3. 7. 11. - - 4n+1 Σ an (say) n=1 (4n-1) (4n+3) (x-3)/2+1 Series, Eland converges if M=1 By Ratio test lim lantil lanl < 1 L 1. at lim lantil 1, test fails. n-p 1 anl Now, lim lantil nto lanl = lim ntoo 37.11. 3.7.11. - =) lim lantil Flim (4n+3)1x-317+1 ntoo lanl 7-7700 4n+1 = lim lantıl n700 Ian' = lim (4n+3) 1x-31 n200 4 (4n-1) (4n+3) 1x-317+1 4n+1 (49-1) 1x-3/m կո x կո 1x-31"
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Follow-up Question

So I worked out the question. And I got the below setup. And I got the interval of convergence as (-1, 7). But when I checked the back of the book, it said the answer was x = 3. Could anyone help with what I did wrong?

3315 T 307-11---Un-1/x-3) we enforce the condition of converg ence of the ratio test fim 3-7-7-3) (x3) ☑ い 。37 11-441-163 14/4731 41 4 un- 4T ⇒ x 344 Interval of convergenc Endpoints TWE 7844 h=| (-1) 3070/1000 (n-1)/4" 14 diverges (3.7-11-0 4 diverges
Transcribed Image Text:3315 T 307-11---Un-1/x-3) we enforce the condition of converg ence of the ratio test fim 3-7-7-3) (x3) ☑ い 。37 11-441-163 14/4731 41 4 un- 4T ⇒ x 344 Interval of convergenc Endpoints TWE 7844 h=| (-1) 3070/1000 (n-1)/4" 14 diverges (3.7-11-0 4 diverges
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Follow-up Question

Why is there an extra (4n - 1) in the top in u(n+1)? 

 

Оли = (+372 3.7.11 (4nes (4n+3) (x-3) 4741 7+1
Transcribed Image Text:Оли = (+372 3.7.11 (4nes (4n+3) (x-3) 4741 7+1
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