# Let $\{a_n\}$ and $\{b_n\}$ be convergent sequences. When is the sequence $a_1, b_1, a_2, b_2, \dots$ convergent and what is its limit?

I'm currently trying to prove that the sequence $$(c_n) = (a_1, b_1, a_2, b_2, \dots)$$ converges only when $$\lim a_n = \lim b_n$$. I know that the sequence will not converge when $$\lim a_n \ne \lim b_n$$. So would I next have to show that $$|c_n-C|<\epsilon$$ and if so how would I choose my epsilon.

• You don't choose the $\epsilon$. It is given. You want to find the $N$. What you want to do for this problem is to write what convergence in $a_n$ and $b_n$ give you and then notice that we can write the distance from $c_n$ and $C$ in terms of what we already got from $a_n$ and $b_n$. Dec 17 '18 at 0:47
• sorry I meant N Dec 17 '18 at 0:49
• I'm a little confused how can rewrite $c_n$ and $C$ in terms of $a_n$ and $b_n$ would it be something like |($a_n$+$b_n$)-($A$+$B$)|=|$c_n$-$C$| Dec 17 '18 at 0:53

You know that there exists a $$N_1$$ and a $$N_2$$ such that the respective distances to the limit are less than $$\epsilon$$. Then you can take $$N:=\mathrm{max}(N_1,N_2)$$ and have the result.
• I understand that but, how does that help me prove that $c_n$ is convergent? Dec 17 '18 at 1:33
• Then you have that $|c_n-C|<\epsilon$ for $n>2N$ where $C$ denotes the limit of both sequences. Dec 17 '18 at 1:48