Let $0<y_{n}$ be a sequence such that $\frac{y_{n+1}}{y_{n}}\to C$for some $0<C<\infty.$ Let $x_{k}=y_{n_{k}}$ be a subsequence of$y_{n}$ such that $\frac{x_{k+1}}{x_{k}}\to B,$ for some $0\leq B\leq\infty.$What are the possible values for $B$ as you chose different subsequence$y_{n_{k}}$?
My feeble attempt: Since $\frac{y_{n+1}}{y_{n}}\to C,$ very roughly (the proof of the ratiotest shows) $y_{n}\approx a\cdot C^{n},$ for some $0<a$ and large$n,$ so $\frac{x_{k+1}}{x_{k}}\approx C^{n_{k+1}-n_{k}},$ for large$k.$ Hence, it appears the set of $B$'s equals the set $\left\{ 0,C,C^{2},C^{3},\ldots\right\},$when $C<1,$ the set $\left\{ 1\right\},$ when $C=1,$ and the set$\left\{ C,C^{2},C^{3},\ldots,\infty\right\},$ when $C>1.$
There is no reason to consider the third case. Indeed, if $C>1$ and$z_{n}:=1/y_{n},$ then $\frac{z_{n+1}}{z_{n}}\to\frac{1}{C}.$
This question was inspired by Comparison principle for order of convergence
Edit: This is not as simple as my original question might have implied. Let $p$ be a real number and let $y_{n}:=\tfrac{1}{n^{p}}$ for $n=1,2,3,\ldots.$Then $\tfrac{y_{n+1}}{y_{n}}=\left(\tfrac{n}{n+1}\right)^{p}\to1.$Let $m>1$ be an integer and let $n_{k}:=m^{k-1},$ then$$\frac{x_{k+1}}{x_{k}}=\left(\frac{m^{k-1}}{m^{k}}\right)^{p}=\frac{1}{m^{p}}.$$Hence, $C=1$ and the set $B$ contains all the values $\tfrac{1}{m^{p}},$$m=2,3,4,\ldots.$