Consider the sequence bnn greaterthanorequalto 1 such that b

Consider the sequence (b_n)_n greaterthanorequalto 1 such that b_n notequalto 0 for all n greaterthanorequalto 1 and lim_n rightarrow infinity b_n notequalto 0. Prove that the sequence (1/b_n)_n greaterthanorequalto 1 is bounded. Assume that the sequence (a_n)_n greaterthanorequalto 1 is convergent. If the sequence (b_n)_n greaterthanorequalto 1 satisfies assumptions of part (a) prove that lim_n rightarrow infinity a_n/b_n = lim_n rightarrow infinity a_n/lim_n rightarrow b_n

Solution

a> From the given conditions, it is clear that limit of the sequence {b_n} exists .

So it is convergent .

Now {b_n} is convergent

=> {1/ b_n } is also convergent

=> {1/b_n } is bounded ( as every convergent sequence is bounded)