If $\{x_{2m}\}$ and $\{x_{2m-1}\}$ converge to the same limit, does $\{x_m\}$ converge? 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 $\{x_m\}$?
 A: 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 $\vert x_m - \lambda^* \vert < \epsilon$.
A: If both subsequences converge to the same limit we surely see that the sequence has to converge. A sequence converges, iff all of it's subsequences converge. Taking an arbitrary subsequence of $(a_n)_{n\in \mathbb{N}}$ we get for every $\varepsilon>0$ some $N_1$ for the odd indices and some $N_2$ for the even such that $$|a_n-a|\leq \varepsilon$$ for all $n\geq \max\{N_1, N_2\}$. 
But the rate of convergence completly gets lost, take $x_{2m}=a$ as the constant sequence of your limit. That surely converges superlinear, take any other superlinear convergent sequence with the same limit now we see that for  $$|x_{k+1}-a| \leq \mu \cdot |x_k -a|^r$$  there surely  aren't $\mu, r$ such that it holds for all $k \in \mathbb{N}$.
Edit: Assuming the sequence to be strictly increasing makes things much easier as the convergence of $x_{2m}$ is already enough to conclude $x_m$  converges. In this case the rate of convergence gets lost too, I mention an idea for a counterexample in the comments. 
