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PF: If given $ \epsilon >0 $, let $ N=1$ so whenever $ n>N $ we have $ |c_n-c|=|c-c|=0 < \epsilon $. Therefore the limit of $ c_n=c $ as $ n \rightarrow \infty $ as required.

This is the answer, but I don't understand why we let $ N=1 $. What would be your solution? Thank you!

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Please try out LateX. It makes things easier for everyone. As for your question, just look at the definition of convergence of a sequence. It should be clear as to why N=1. What you have is a constant sequence. –  Vishesh Oct 2 '12 at 4:27
I TeX'd the title question... you can take a look at the markup I used by clicking the edit timestamp. It's very straightforward, so feel free to give it a try in the future! –  rschwieb Oct 2 '12 at 12:41

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up vote 4 down vote accepted

The idea is that if you claim that $\lim_{n \to \infty} c_n=c$, you have to be prepared for me to challenge you with an $\epsilon$, known to be greater than $0$. If I do, you have to respond with an $N$ that says for every $n \ge N$, $c_n$ is within $\epsilon$ of $c$. That means that eventually it gets very close to $c$ and stays there. In this case, we are already there, so $N=1$ works, but so would any higher $N$.

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The convergence of a sequence to ,say a point $ x $ is sort of equivalent to saying that no matter how small a distance you choose(distance being $ \epsilon $ ), there will always be elements of the sequence within that prescribed distance from $ x $. The $ N $ here is that juncture of the sequence after which the above criterion is satisfied. Its clear that its determined by the distance you choose. The smaller the distance you choose, greater the likelihood that your $ N $ will be large(this is also determined by the sequence in consideration). In your case, we have a constant sequence, so the elements are at a constant distance from one another, so $ N=1 $ will suffice.

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Thank you for awesome explanation! –  Maximiliano Oct 2 '12 at 4:54

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