Question #340457

) Give an example to show that if the convergence of 􏰄 an is conditional and (bn) is a bounded

sequence, then 􏰄 anbn may diverge.


1
Expert's answer
2022-05-16T12:25:36-0400

ANSWER :n=1an=n=1(1)nn,bn=(1)n\sum_{n=1}^{\infty}a_{n}=\sum_ {n=1}^{\infty}\frac{(-1)^{n } }{n},\, \, b_{n}=(-1)^{n}

EXPLANATION The series n=1(1)nn\sum_ {n=1}^{\infty}\frac{(-1)^{n } }{n} is an alternating series, since the sequence an=(1)ncna_{n}=(-1)^{n}c_{n} and cn=1nc_{n}= \frac {1}{n } is decreasing , limncn=0\lim _{n\rightarrow\infty}c_{n}=0 . The series n=11n\sum_ {n=1}^{\infty}\frac{1 }{n} diverges, because it is a pp- series for p=1p=1 , hence the series n=1(1)nn\sum_ {n=1}^{\infty}\frac{(-1)^{n } }{n} converges conditionally . The sequence bn=(1)nb_{n}=(-1)^{n} is a bounded sequence , since 1bn1-1\leq b_{n}\leq 1 for all nNn\in\N (b2n=1,b2n1=1b_{2n }=1,\, b_{2n-1}=-1 ). Since anbn=1na_{n}\cdot b_{n}=\frac {1}{n} , then the series n=1anbn=n=11n\sum_{n=1}^{\infty}a_{n}\cdot b_{n}=\sum_ {n=1}^{\infty}\frac{1 }{n} diverges.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS