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Question Number 676 by 123456 last updated on 22/Feb/15

g i v e n t w o s e q u e n c e a_{n} > 0, b_{n} > 0 s u c h

\forall n \in N^{*}, a_{n}^{n} < b_{n} < a_{n}^{1 / n}

a . i f \underset{n = 1}{\sum^{+ \infty}} a_{n} c o n v e r g e t h e n \underset{n = 1}{\sum^{+ \infty}} b_{n} c o n v e r g e ?

b . p r o o f t h a t i f a_{n} \in (0, 1) t h e n b_{n} \in (0, 1)

c . (d i s) p r o o f t h a t i f a_{n} \to 1 t h e n b_{n} \to 1

Commented by prakash jain last updated on 22/Feb/15

b . 0 < a_{n} < 1 \Rightarrow 0 < a_{n}^{n} < 1

0 < a_{n} < 1 \Rightarrow 0 < a_{n}^{1 / n} < 1

a_{n}^{n} < b_{n} < a_{n}^{1 / n} \Rightarrow 0 < b_{n} < 1

c . ∵ a_{n} \to 1 \Rightarrow a_{n}^{n} \to 1 and a_{n}^{1 / n} \to 1

∴ a_{n} \to 1 \Rightarrow b_{n} \to 1

a . If a_{n} converges we can say a_{n}^{n} converges .

but we cannot conclude anything about

a_{n}^{1 / n} . Hence we cannot conclude that b_{n}

onverges .