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Question Number 657 by 123456 last updated on 21/Feb/15

i f (a_{n}) a n d (b_{n}) a r e t w o r e a l s e q u e n c e

s u c h t h a t e^{a_{n}} = a_{n} + e^{b_{n}}

a) p r o o f t h a t a_{n} > 0 \Rightarrow b_{n} > 0

b) i f a_{n} > 0 \forall n \in N i f \underset{n = 0}{\sum^{+ \infty}} a_{n} c o n v e r g e t h e n

\underset{n = 0}{\sum^{+ \infty}} \frac{b_{n}}{a_{n}} c o n v e r g e

Answered by prakash jain last updated on 20/Feb/15

e^{a_{n}} = 1 + a_{n} + \frac{a_{n}^{2}}{2!} + \dots

a_{n} > 0 \Rightarrow e^{b_{n}} = e^{a_{n}} - a_{n} > 1 \Rightarrow b_{n} > 0