Find the exact interval of convergence for the series in Example 6. Tell clearly using what method when it converges and and what method if diverges when checking for the interval. Example 6=0Consider 2-³n. This is deceptive, and it is tempting to cal-culate 3 = lim sup(2-n)1/n = and conclude R=2. This is wrongbecause 2-n is the coefficient of x³ not xn, and the calculation ofß must involve the coefficients an of xn. We need to handle thisseries more carefully. The series can be written ªnx" where2-k and an = 0 if n is not a multiple of 3. We calculateby using the subsequence of all nonzero terms, i.e., the subsequencegiven by o(k)= 3k. This yields∞in=0azk =∞B = lim sup |an|¹/nВ= lim |a3k|¹/3k = lim (2-k)¹/³k = 2−1/3k→∞k→∞1Therefore the radius of convergence is R = = 2¹/3.

Answered: Image /qna-images/answer/c230ecf9-f082-4708-a9de-9a88f3b57d23.jpg

8 Consider 2-³. This is deceptive, and it is tempting to cal- n=0 = culate B lim sup(2-¹)¹/n = and conclude R 2. This is wrong because 2-n is the coefficient of x³ not xª, and the calculation of B must involve the coefficients an of x. We need to handle this series more carefully. The series can be written Σ0 anx" where azk 2-k and an = = 0 if n is not a multiple of 3. We calculate ß by using the subsequence of all nonzero terms, i.e., the subsequence given by o(k)= 3k. This yields in=0 = 3 lim sup an | 1/n = lim |a3k|1/3k = k→∞ = _lim (2−k)¹/³k = 2−¹/3 k→∞ Therefore the radius of convergence is R = = = 2¹/3

8 Consider 2-³. This is deceptive, and it is tempting to cal- n=0 = culate B lim sup(2-¹)¹/n = and conclude R 2. This is wrong because 2-n is the coefficient of x³ not xª, and the calculation of B must involve the coefficients an of x. We need to handle this series more carefully. The series can be written Σ0 anx" where azk 2-k and an = = 0 if n is not a multiple of 3. We calculate ß by using the subsequence of all nonzero terms, i.e., the subsequence given by o(k)= 3k. This yields in=0 = 3 lim sup an | 1/n = lim |a3k|1/3k = k→∞ = _lim (2−k)¹/³k = 2−¹/3 k→∞ Therefore the radius of convergence is R = = = 2¹/3

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

Find the exact interval of convergence for the series in Example 6. Tell clearly using what method when it converges and and what method if diverges when checking for the interval.

Example 6
=0
Consider 2-³n. This is deceptive, and it is tempting to cal-
culate 3 = lim sup(2-n)1/n = and conclude R=2. This is wrong
because 2-n is the coefficient of x³ not xn, and the calculation of
ß must involve the coefficients an of xn. We need to handle this
series more carefully. The series can be written ªnx" where
2-k and an = 0 if n is not a multiple of 3. We calculate
by using the subsequence of all nonzero terms, i.e., the subsequence
given by o(k)= 3k. This yields
∞
in=0
azk =
∞
B = lim sup |an|¹/n
В
= lim |a3k|¹/3k = lim (2-k)¹/³k = 2−1/3
k→∞
k→∞
1
Therefore the radius of convergence is R = = 2¹/3.

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