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I am considering a following problem: Does $$\\a_{2k}, a_{2k+1}, a_{3k} \rightarrow g $$ imply that $$a_{n} \rightarrow g ?$$ I know that if every subsequence goes to $g$ then also a sequence goes to $g$. My way of reasoning is following: we know that odd and even $k$ subsequences goes to $g$. But if we substitute for example $k=t^{2} -5$, then there is problem (at least for me). If I can reason that way, what must I do to prove more formally that the statement is false? And if I can not, what is going on with $a_{n}$? Thanks for any hints!

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If the even numbered terms and odd numbered terms both converge to the same limit $L$, then the limit of the sequence exists and is $L$.

To prove this, let $\epsilon > 0$ be arbitrary. We want to find $N \in \mathbb{N}$ such that $n > N$ implies $|a_n - L| < \epsilon$.

Since the even numbered terms are all within epsilon of $L$ after some $N_1 = 2j$ and the odd numbered terms are all within epsilon of $L$ after some $N_2 = 2k + 1$, just pick $N = \max\{N_1, N_2\}$.

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You may want to prove the next nice lemma:

Lemma: If $\,\{A_i\}_{i\in I},$ is some partition of $\,\Bbb N\,$ s.t. $\,|A_i|=\aleph_0=|\Bbb N|\,\,\,,\,\forall\,\,i\in I\,$ , and all the sets $\,A_i\,$ are well ordered, then for a real sequence $\,\{x_n\}\,$ it is true that

$$x_n\xrightarrow [n\to\infty]{} x\Longleftrightarrow (\,\forall\,\,i\in I\,\,,x_{n_i}\xrightarrow [n_i\to\infty\,,\,n_i\in A_i]{} x)$$

Thus, as B. wrote in his answer, it is enough and sufficient that the subsequence of odd indexes and the one of even indexes converge both to the same limit.

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  • $\begingroup$ Doesn't $I$ have to be a finite in order for this to work? $\endgroup$ – user642796 Nov 1 '12 at 17:48
  • $\begingroup$ Well, not that you mention it...I'm not completely sure, I'll try to check this later. Thanks. $\endgroup$ – DonAntonio Nov 1 '12 at 18:23
  • $\begingroup$ Take $I = \{ 1 \} \cup \{ p : p \text{ is prime} \}$. Define $A_1 = \{ 1 \} \cup \{ n : n \text{ is not a prime power} \}$, and $A_p = \{ p^k : k \geq 1 \}$ for prime $p$. Define the sequence $( x_n )_{n \geq 1}$ so that $x_n = 0$ for all $n \in A_1$ and $x_{p^k} = \frac{1}{k}$ for $p$ prime and $k \geq 1$. Then for each $i \in I$ the sequence $( a_n )_{n \in A_i}$ converges to $0$, but the sequence $( a_n )_{n \geq 1}$ does not converge as it takes the value $1$ infinitely often and has subsequences converging to $0$. $\endgroup$ – user642796 Nov 1 '12 at 18:39
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In the other answers you have seen that it $a_{2k},a_{2k+1}\to g$ then already $a_n\to g$. You don't even need $a_{3k}$. I would like to add that if you just require $a_{2k},a_{2k+1}$ and $a_{3k}$ to converge (a priori with different limits), then you have that all three limits coincide and that $a_n$ has the same limit.

To see this note that $a_{2n}$ and $a_{3n}$ as well as $a_{2n+1}$ and $a_{3n}$ have a commen subsequence. Then you only need that if a converging sequence has a subsequence with limit $g$ then the whole sequence has limit $g$.

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