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.

  • 2
    $\begingroup$ 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 $b_n$. $\endgroup$
    – twnly
    Dec 17 '18 at 0:47
  • $\begingroup$ sorry I meant N $\endgroup$ Dec 17 '18 at 0:49
  • $\begingroup$ 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 |($a_n$+$b_n$)-($A$+$B$)|=|$c_n$-$C$| $\endgroup$ 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.

  • $\begingroup$ I understand that but, how does that help me prove that $c_n$ is convergent? $\endgroup$ Dec 17 '18 at 1:33
  • $\begingroup$ Then you have that $|c_n-C|<\epsilon$ for $n>2N$ where $C$ denotes the limit of both sequences. $\endgroup$
    – sehigle
    Dec 17 '18 at 1:48

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