Find relation between these two subsequences

Question

If$(x_n)$ be a sequence of real number such that subsequences $(x_{2n})$ and $(x_{3n})$ converges to K and L respectively.Then what is right-

$\hspace{.5em}\begin{array}{rl} a. & (x_n) \hspace{.5em} always \hspace{.5em} converges \\b. & if \hspace{.5em} K =L \hspace{.5em} then \hspace{.5em} (x_n) \hspace{.5em} converges  \\ c. & (x_n) \hspace{.5em} maynot  \hspace{.5em} converge \hspace{.5em} but \hspace{.5em} K=L \\ d. & it \hspace{.5em} is \hspace{.5em} possible \hspace{.5em} to \hspace{.5em} have \hspace{.5em} K \neq L \end{array}$

I think wrong options can be eliminated by choosing a proper sequence but how to find it?

Answer 1

Under those hypothesis, the sequence may not converge, but $K=L$. That's so because $(x_{6n})_{n\in\mathbb N}$ is a subsequence of both subsequences and therefore it converges to both $K$ and $L$.

However if you define$$x_n=\begin{cases}1&\text{ if $n$ is prime and }n>3\\0&\text{ otherwise,}\end{cases}$$then $(x_n)_{n\in\mathbb N}$ devierges, but both sequences $(x_{2n})_{n\in\mathbb N}$ and $(x_{3n})_{n\in\mathbb N}$ converge to $0$.

Answer 2

Correct me if wrong :

a) False.

b) False.

c) True.

d) $K=L.$

a)Consider the subsequence $x_p =p$ ; where $p$ is prime.

This subsequence does not converge, hence $x_n$ does not converge.

b) False, refer to argument in a).

c)True. $x_{2n}$ and $x_{3n}$ have a common subsequence $x_{6n}$, hence $K=L.$

d) $K=L$, argument in c).