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Given a sequence $a_n$, I know that if I can find a divergent subsequence of $a_n$, or two subsequences of $a_n$ that converge to different values, $a_n$ diverges, since, if I have understood correctly, a sequence $a_n$ converges to a limit $L$ if and only if every subsequence of $a_n$ converges to that value $L$.

I've been wondering if this last condition was equivalent to showing that some subsequences converge to $L$, picking the subsequences such that every element of the original sequence is in at least one of the subsequences. Is it? I would guess that the terms "partition" or "covering" fit this description.

Thanks.

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No. Let a_n alternate between 1 and -1; then the even and odd subsequences have the desired property but a_n does not converge. –  Qiaochu Yuan Feb 9 '11 at 22:21
    
@Qiaochu: That doesn't quite work, because those subsequences have different limits. –  Jonas Meyer Feb 9 '11 at 22:25
    
@Jonas: ah, I misread the question. –  Qiaochu Yuan Feb 9 '11 at 22:42
    
Sorry, what is the question? I read it twice and am still not sure. Call me a stickler for punctuation, but I was really hoping to encounter a helpfully placed question mark. –  Pete L. Clark Feb 10 '11 at 2:01
    
I've edited half-jokingly to include a question mark. –  Abel Feb 10 '11 at 2:05

2 Answers 2

up vote 2 down vote accepted

If you can partition a sequence into finitely many subsequences, each of which converges to $L$, then the original sequence must converge to $L$ as well. This is clear from the following definition of the limit: $a_n \rightarrow L$ iff for all $\epsilon>0$ there exists $N_\epsilon$ such that $|a_n - L| < \epsilon$ whenever $n \ge N_\epsilon$. But then for any $\epsilon > 0$, because the $i$-th subsequence converges to $L$, it is within $\epsilon$ of the limit for $n \ge N^{(i)}_\epsilon$; so the sequence itself is within $\epsilon$ of the limit for $n \ge N_\epsilon = \max_{i}N^{(i)}_\epsilon$.

However, the result does not hold for a partition into infinitely many subsequences. For instance, consider the case where $a_n=1$ when $n$ is prime and $a_n=0$ otherwise. This can be partitioned into infinitely many subsequences, where $a_n$ is in the $i$-th subsequence if the smallest prime factor of $n$ is the $i$-th prime. (Just put $a_1$ anywhere.) Each subsequence converges (immediately) to zero, but the original sequence does not converge, because it has sporadic $1$'s as far out as you care to look.

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Nice example! Mine only has the weaker covering condition. –  Jonas Meyer Feb 9 '11 at 22:45

Consider the sequence $(x_n)_n$ such that $x_n=0$ if $n$ is even and $x_n=1$ if $n$ is odd. Let $m$ be a positive integer, and consider the subsequence $(x_{n_k})_k$ such that $x_{n_k}=x_k$ if $k\leq m$, and $x_{n_k}=x_{2k}$ if $k\gt m$. Now consider the collection of such subsequences as $m$ varies over the positive integers. They all are eventually constantly $0$, and they "cover" the original sequence. An analogous covering construction could be given for any sequence with a convergent subsequence.

Related to your remarks before the question, it is often useful to use the criterion that a sequence converges to $L$ if (and only if) every subsequence has a further subsequence that converges to $L$.

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