Given a sequence $x_1<x_2<x_3<x_4<\ldots,$ I found both subsequences $\{x_{2m}\}$ and $\{x_{2m-1}\}$ converge to $\lambda^*$. In this case, can I say $\{x_m\}$ also converge to $\lambda^*$? (i.e. a sequence $\{x_m\}$ is divided into two subsequences $\{x_{2m}\}$ and $\{x_{2m-1}\}$)
If so, if the rate of convergence is superlinear for both subsequences $\{x_{2m}\}$ and $\{x_{2m-1}\}$, can I also say that $\{ x_m\}$ converge superlinearly?
What is more curious to me is the case where two subsequnces converge at different rates of convergence. For example, if $\{x_{2m}\}$ converges linearly to $\lambda^*$ and $\{x_{2m-1}\}$ converges superlinearly to $\lambda^*$, what is the rate of convergence of $\{x_m\}$?
Best Answer
Yes, you can conclude that the sequence converges. This is because, given $\epsilon>0$, there exists $N_1(\epsilon)$ such that for all $2m>N_1(\epsilon)$, $\vert x_{2m} - \lambda^* \vert < \epsilon$. Similarly, given $\epsilon>0$, there exists $N_2(\epsilon)$ such that for all $2m>N_2(\epsilon)$, $\vert x_{2m+1} - \lambda^* \vert < \epsilon$. Hence, choosing $N(\epsilon) = \max \{N_1(\epsilon), N_2(\epsilon)\}$, we can conclude that for all $m > N(\epsilon)$, we have $\vert x_m - \lambda^* \vert < \epsilon$.