# Convergent Subsequence in $\mathbb R$

Let $\langle a_n \rangle = \dfrac{(-1)^n}{1+n}$ be a sequence in $\mathbb R$.

Considering the limit point(s) of this sequence and the subsequences that converge to this point, I have two subsequences: $$a_{2k} = \frac{1}{1+2k} \to 0 \text{ and } a_{2k+1} = \frac{-1}{1+2k+1} \to 0$$

If the question asks for the limit points of the sequence, and a subsequence that converges to this limit point, do I leave out the second subsequence? I don't think I've missed a limit point, but it doesn't hurt to check.

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## 1 Answer

If you want to find the limit points, you have to think about the subsequential limits for every subsequence, not just those two. But luckily, the original sequence converges (to 0) and in a convergent sequence, every subsequence converges to the same limit as the whole sequence does.

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Is it fair to say that since those subsequences are the only ones I have to consider because they are the only two convergent subsequences? Other than ones like $\langle a_{4k} \rangle$ which are contained within the one considered? –  KingOliver Aug 14 '12 at 21:41
You can take a subsequence that includes infinitely many terms from each of those subsequences and it will still converge to 0. Like I said, $\textbf{every}$ subsequence will converge to 0, not just ones that have only even terms or only odd terms. –  Francis Adams Aug 14 '12 at 21:45
But if you just want the limit points, the answer is just 0, even if it comes from more than one subsequence. –  Francis Adams Aug 14 '12 at 21:50
Never mind, of course you are right, because the denominator $\to \infty$ so the sequence $\to 0$, sorry I lost perspective. Thanks again! –  KingOliver Aug 14 '12 at 21:51