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.


Expert's answer

ANSWER :∑n=1∞an=∑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 , lim⁡n→∞cn=0\lim _{n\rightarrow\infty}c_{n}=0 . The series ∑n=1∞1n\sum_ {n=1}^{\infty}\frac{1 }{n} diverges, because it is a p−p- 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 −1≤bn≤1-1\leq b_{n}\leq 1 for all n∈Nn\in\N (b2n=1, b2n−1=−1b_{2n }=1,\, b_{2n-1}=-1 ). Since an⋅bn=1na_{n}\cdot b_{n}=\frac {1}{n} , then the series ∑n=1∞an⋅bn=∑n=1∞1n\sum_{n=1}^{\infty}a_{n}\cdot b_{n}=\sum_ {n=1}^{\infty}\frac{1 }{n} diverges.


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