Thread: Limit of sequence {b_n}/n --> b

This should be easy, but I missed a week of my real analysis class and the problem recommends using a theorem not in the text.



Let {b_n} be a sequence in IR with the property that b_n+1 − b_n → b as
n → ∞. Prove that b_n/n → b as n → ∞. [Hint: Use Theorem 3 from
class].

Originally Posted by mickyduu

This should be easy, but I missed a week of my real analysis class and the problem recommends using a theorem not in the text.
Let {b_n} be a sequence in IR with the property that b_n+1 − b_n → b as
n → ∞. Prove that b_n/n → b as n → ∞. [Hint: Use Theorem 3 from
class].

Shame on you for missing class. It would be nice to know what is th. 3.
There is a theorem of means:
If then
Using that we can define (if necessary define ).
Now clearly so .
But .
Because you missed class and because I have no idea about Th3, that proof may not work.

Thanks for the help, but I don't think that your answer works, or more precisely I don't think that it answers the question. I'm pretty sure that I need to show that the


b_n/n → b as n → ∞

where the b_n is a term of the ORIGINAL {b_n} sequence.

My guess is that I'm going to need to start with the definition of a limit for a convergent sequence:

"A real number x is the limit of the sequence (x_n) if the following condition holds:for each ε > 0, there exists a natural number N such that, for every , we have .In other words, for every measure of closeness ε, the sequence's terms are eventually that close to the limit. The sequence (x_n) is said to converge to or tend to the limit x, written or."


I believe that theorem 3 is going to be a common analysis theorem, possibly one that is better known in some generalized form or when dealing with functions2nd +, however our text merely glosses over dealing with sequences.

Originally Posted by mickyduu

Thanks for the help, but I don't think that your answer works, or more precisely I don't think that it answers the question. I'm pretty sure that I need to show that the
b_n/n → b as n → ∞

Yes it does. PLease see my EDIT. There was a cut&paste error.

It proves that so