Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.


share|improve this question
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.

share|improve this answer
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 $L$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.