# Sequences

• May 2nd 2010, 05:09 AM
Webby
Sequences
Hi I am having a bit of trouble expressing the answer to a problem. If it is any constellation this problem goes alongside convergence proofs.

$\displaystyle Let \{a_n \}^\infty_1 , \{b_n\}^\infty_1$ be sequences such that $\displaystyle lima_n = a$ and $\displaystyle b_n = a_n$ if n is even
or $\displaystyle b_n = a$ if n is odd. Show that $\displaystyle limb_n = a$

That being said if n is even $\displaystyle b_{2n} = a_2 + a_4 + ... + a_{2n}$ . As $\displaystyle a_{2n}$ is a subsequense of $\displaystyle a_n$ one can conclude that $\displaystyle a_{2n}$ converges to $\displaystyle a$. Hence $\displaystyle b_n$ will converge to a if n is even.
As $\displaystyle b_{2n+1} = a$ is a is a constant. $\displaystyle b_{2n+1}$ will also converge to $\displaystyle a$.

As $\displaystyle b_n$ converges to a in both cases, then $\displaystyle limb_n = a$ as stated.

I think this is the idea behind the question, if I overlooked anything please tell me. Thank you. (Happy)
• May 2nd 2010, 05:44 AM
Tikoloshe
Quote:

Originally Posted by Webby
Hi I am having a bit of trouble expressing the answer to a problem. If it is any constellation this problem goes alongside convergence proofs.

You mean consolation.
Quote:

Originally Posted by Webby
As $\displaystyle b_n$ converges to a in both cases, then $\displaystyle limb_n = a$ as stated.

I think this is the idea behind the question, if I overlooked anything please tell me. Thank you. (Happy)

If you want to prove convergence of a sequence by subsequences, you have to show that every subsequence (not just the two you have selected) converges to the same number. An alternate route is to show that given $\displaystyle \epsilon >0, \exists N$ such that $\displaystyle \forall n\geq N$, you have$\displaystyle \lvert b_n-a\rvert < \epsilon$. You can find $\displaystyle N$ from the $\displaystyle N$ you would need for $\displaystyle a_n$. Thus $\displaystyle \lim b_n=a$.
• May 2nd 2010, 06:05 AM
Webby
This may sound like a silly question but how would you find n>N in this case $\displaystyle \lvert a_n-a\rvert < \epsilon$ where $\displaystyle n > \epsilon$.
• May 2nd 2010, 06:15 AM
Defunkt
Quote:

Originally Posted by Webby
This may sound like a silly question but how would you find n>N in this case $\displaystyle \lvert a_n-a\rvert < \epsilon$ where $\displaystyle n > \epsilon$.

Since $\displaystyle a_n \to a$, for any $\displaystyle \epsilon >0$ there exists $\displaystyle n_0 \in \mathbb{N}$ such that $\displaystyle n>n_0 \Rightarrow|a_n-a|<\epsilon$. You need to show that the same $\displaystyle n_0$ works for the sequence $\displaystyle b_n$.
• May 2nd 2010, 09:42 AM
alexandrabel90
can i say that let N = max{ 2no, 2no +1 } then for all n> N, bn converges to a?