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Where $b_1 = 2$. I've been attemting to prove it using the $\epsilon-M$ definition of limit but I can't find a value of M. That is, I've been trying to prove: $\forall \epsilon > 0\ \exists M \in \mathbb{N}\ \forall n > M\ |b_{2n}-b_{2n-1}| < \epsilon$. All I've managed to think to do is try to bound the expression $|1+\dfrac{1}{b_{2n-1}}-b_{2n-1}|$ to see if it might help but I've been unsuccesful. Even if I did bound it, I cannot find a value of M since I cannot solve for n. Edit: currently, all I can use is that 1. $(b_{2n})$ and $(b_{2n-1})$ converge and 2. $(b_{2n})$ is monotonic increasing while $(b_{2n-1})$ is monotonic decreasing.

StubbornAtom
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LASER BEAR ASSAULT UNIT
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  • Compare https://math.stackexchange.com/q/2205883/42969. – Martin R Sep 25 '19 at 13:32
  • My goal is not to show that $b_n$ converges. It is to show that the subsequences $b_{2n}$ and $b_{2n-1}$ converge to the same limit. This seems to be a different aim than the question linked. I cannot prove that the $(b_n)$ converges first and then conclude that the two subsequences must converge to the same value either. – LASER BEAR ASSAULT UNIT Sep 25 '19 at 13:37

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