Using the definition of a limit to prove 1/n converges to zero.

Question

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?

Answer 1

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.

Answer 2

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. $n \geq N \Rightarrow \frac{1}{n} \leq \frac{1}{N} \lt \epsilon$

Answer 3

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.

Answer 4

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 $0$.

Answer 5

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!

Answer 6

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.