Sequences

**Webby** · May 2nd 2010, 05:09 AM

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.

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

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

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

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

**Tikoloshe** · May 2nd 2010, 05:44 AM

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.

Originally Posted by Webby

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

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

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 $\epsilon >0, \exists N$ such that $\forall n\geq N$ , you have $\lvert b_n-a\rvert < \epsilon$ . You can find $N$ from the $N$ you would need for $a_n$ . Thus $\lim b_n=a$ .

**Webby** · May 2nd 2010, 06:05 AM

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

**Defunkt** · May 2nd 2010, 06:15 AM

Originally Posted by Webby

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

Since $a_n \to a$ , for any $\epsilon >0$ there exists $n_0 \in \mathbb{N}$ such that $n>n_0 \Rightarrow|a_n-a|<\epsilon$ . You need to show that the same $n_0$ works for the sequence $b_n$ .

**alexandrabel90** · May 2nd 2010, 09:42 AM

can i say that let N = max{ 2no, 2no +1 } then for all n> N, bn converges to a?

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