Answered: n = 1 18 (−1)”+ ¹ 3 · 7 · 11 · · · (4n…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

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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=

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Follow-up Questions

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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"

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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?