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So we define a sequence as a sequence ${a_n}$ is said to converge to a number $\alpha$ provided that for every positive number $\epsilon$ there is a natural number N such that |${a_n}$ - $\alpha$| < $\epsilon$ for all integers n $\geq$ N.

What I'm not understanding is what does this mean. For example, $\frac{1}{n}$ converges to 0. But I don't understand how I use this definition to prove that this converges to 0. It sounds trivial but how do I use the definition to prove that $\frac{1}{n}$ converges to 0. Can you also show the reasoning as to why you use certain steps?

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    $\begingroup$ Seems to me the title should rather start with 'Using the definition of a limit to prove...' $\endgroup$
    – CiaPan
    Aug 6, 2015 at 13:46

6 Answers 6

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Let's try and fit your definition into the example you mentioned, first. The sequence $a_n$ you gave is $a_n = \frac{1}{n}$, and the limit $\alpha$ is $\alpha = 0$. Therefore, we wish to prove that for any $\varepsilon > 0$ there is a positive integer $N$ such that if $n \geq N$, then $|\frac{1}{n} - 0| = \frac{1}{n} < \varepsilon$.

Let's think about that definition for a moment. What this says is that eventually, every term of the sequence $\frac{1}{n}$ is close to $0$, no matter how arbitrarily close we want to be. And really, that's all we mean by convergence: eventually, the terms of the sequence get "close" to the limit. We are just making that notion of closeness precise.

Now, let's prove the result. Let $\varepsilon > 0$ be given. Then there is a positive integer N such that $\frac{1}{N} < \varepsilon$ (this is the Archimedean Property). Of course, when $n \geq N$, we have that $\frac{1}{n} \leq \frac{1}{N}$ by dividing both sides by $n$ and $N$. This same procedure works for any $\varepsilon$; there is nothing special here about the one we chose (though $N$ might be different in each case; that's not a problem). Therefore, given any $\varepsilon > 0$, we can find a positive integer $N$ such that for $n \geq N$, $|\frac{1}{n} - 0| < \varepsilon$. That is, we showed that $a_n = \frac{1}{n}$ converges to $0$ by definition, as desired.

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  • $\begingroup$ I'm confused as to why it works for any $\epsilon$. Because my real analysis book says that it should work for every $\epsilon$ > 0. Can I choose my $\epsilon$ or is it like any value? $\endgroup$ Aug 6, 2015 at 13:58
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    $\begingroup$ The epsilon you choose is arbitrary. What is significant here is that the same general procedure works, regardless of any specific epsilon. However, it is crucial that the definition can be satisfied for any epsilon; simply picking one is not enough (consider, for example, $\varepsilon = 2$, $a_n = \pm 1$ assigned randomly, and $\alpha = 0$. Certainly every $a_n$ is within $\varepsilon$ of $0$, but there is no way that this random sequence converges to anything.) $\endgroup$ Aug 6, 2015 at 14:05
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Often, it helps to work backwards. So for this, you must choose $N$ such that $|\frac{1}{n} - 0| = \frac{1}{n} \lt \epsilon$ whenever $n \geq N$ for any fixed $\epsilon$. Shuffling the first inequality around says $\frac{1}{\epsilon} \lt n$. So, simply pick $N \gt \frac{1}{\epsilon}$. This gives you the result, i.e. $n \geq N \Rightarrow \frac{1}{n} \leq \frac{1}{N} \lt \epsilon$

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    $\begingroup$ N has to be an element of the natural numbers because it is an index. You should make use of the rounding function [...]. [16.2] = 17. $\endgroup$
    – jubueche
    Jun 8, 2017 at 17:33
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Stated loosely, the definition reads as: "The sequence $\{a_n\}$ converges to $a$ if terms of the sequence get arbitrarily close to $a$ in absolute value."

So, how close is "close"?

Give me any positive distance (this is your $\varepsilon > 0$). If $a_{n} \to a$, then after some point in the sequence (this is your $N$), all of the terms $a_{n}$ should be close (within $\varepsilon$) to $a$.

Now: if you want to show $\frac1n \to 0$, then you should be able to do this. How close do you want to be to $0$? Let's say you want to be "precise": you want to get within $0.0001$ of $0$. But you know $0.0001 = \frac{1}{10000}$, and so any term further along in the sequence than $a_{10000}$ will be closer to $0$ than $0.0001$. But you can always do better! For any arbitrarily small number $\varepsilon >0$, just make sure $N$ is larger than $\frac{1}{\varepsilon}$!

This was the more informal discussion of the argument -- the formal solution will more closely resemble what others have posted.

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Let $\epsilon>0$ be given.

\begin{equation} \left|\frac{1}{n}-0\right|=\left|\frac{1}{n}\right|=\frac{1}{n}~~~~~~~~~~~~~~~~~~(1) \end{equation}

By Archimedean property for $ϵ>0$, there exists an $n_0\in \mathbb{N}$ such that \begin{equation} n_0 ϵ>1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2) \end{equation}

Now, for all $n≥n_0$, the eq. (1) becomes

\begin{equation} \left|\frac{1}{n}-0\right|=\frac{1}{n}\leq\frac{1}{n_0}<\epsilon. \end{equation}

This proves that the sequence $\{1/n\}$ converges to the limit $0$.

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  • $\begingroup$ Your answer adds nothing new to the already existing answers. $\endgroup$ Jun 23, 2019 at 11:22
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Suppose you have chosen your $\epsilon$. Then we want to argue that there is some (large) integer $N$ such that $n>N\Rightarrow 0<\frac 1n<\epsilon$.

If you want to look at this abstractly, this follows from the Archimedean property for the real line (https://en.wikipedia.org/wiki/Archimedean_property). This just says that, given any real number $X$ there is a natural number larger than it. (your desired result follows by applying this principle to $\frac {1}{\epsilon}$).

If you want to be more hands on with it, suppose you have written out a decimal expansion for $\epsilon$. Let $M$ denote the place of the first non-zero entry (to the right of the decimal). Then, clearly, $\frac {1}{10^{M+1}} < \epsilon$ so we can choose $N = 10^{M+1}$.

Hope that helps!

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Frankly speaking, whatever tiny neighborhood of $\alpha$ you choose, almost all terms of the $(a_n)$ sequence (that is all, except possibly finitely many) fall inside that neighborhood.

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