# Let $\{x_n\}$ be a sequence such that $\forall k > 1 \in \Bbb N$ its subsequence $\{x_{pk}\}$ converges to $1$. Does $\{x_n\}$ converge to $1$?

Let $$\{x_n\}$$ be a sequence such that $$\forall k > 1 \in \Bbb N$$ its subsequence $$\{x_{pk}\}$$ converges to $$1$$.

Does $$\{x_n\}$$ converge to $$1$$?

I'm a bit confused by this problem since no constraints on $$p$$ are given. The only reasonable way seems to consider $$p$$ a natural number. Intuitively the statement seems to be true. Let's try to consider different values for $$k$$: $$k = 2:\{x_{pk}\} = \{x_2, x_4, \dots, x_{2p} \}\\ k = 3:\{x_{pk}\} = \{x_3, x_4, \dots, x_{3p} \}\\ \cdots\\ k = n:\{x_{pk}\} = \{x_n, x_{2n}, \dots, x_{np} \}\\$$

All of the sequences above converge to $$1$$, namely: $$\lim_{p\to\infty}x_{2p} = 1\\ \lim_{p\to\infty}x_{3p} = 1\\ \cdots\\ \lim_{p\to\infty}x_{np} = 1\\$$

At this point, I was thinking about the union of all the sequences above. All of them are subsequences of $$x_n$$, and since $$k$$ is an arbitrary natural number we may consider the following set: $$X = \bigcup\limits_{k=1}^{\infty} x_{pk}$$

Moreover, since every sequence in the union converges to $$1$$ then it must follow that $$X$$ also converges to one. But $$X$$ is essentially equivalent to $$\{x_n\}$$ which would imply that: $$\lim_{n\to\infty}x_n = 1$$

I'm not sure the reasoning above might be applied to the problem. So I would like to ask for verification or a valid solution in case mine is wrong. Also, in case the idea in mine is fine then is it rigorous enough to consider the proof complete? Thank you!

Of course not (in general). Take $$x_n$$ to be $$0$$ if $$n$$ is prime and $$1$$ otherwise.
• Thank you for your answer. Sorry for the confusion but I've missed adding an important notice about $x_{pk}$. Which is $\{x_{pk}\}$ is a subsequence of $\{x_n\}$. – roman Mar 15 at 17:30
• You are right, after thinking more carefully on this $x_n$ is indeed not necessarily convergent, even given the fact it has convergent subsequences. – roman Mar 15 at 17:51
• $x_n$ is not convergent – mathworker21 Mar 15 at 18:23