To prove : $\lim_{n \to +\infty} a_{n} = 0$, where $a_{n}= \frac{1}{n}$ from definition

Question

We have to prove : $\lim_{n \to +\infty} a_{n} = 0$, where $a_{n}= \frac{1}{n}$.

The proof goes like this : $\forall\epsilon >0 \exists N=N(\epsilon)\in \mathbb{N} : \forall n \geq N \implies \mid \frac{1}{n} - 0 \mid < \epsilon$, from this we get : $n >\frac{1}{\epsilon}$, and from the Archimedian Principle of the Real Numbers we choose : $N=[\frac{1}{\epsilon}]+1 > \frac{1}{\epsilon}$, that's why we get : $\mid \frac{1}{n} - 0 \mid = \frac{1}{n} \leq \frac{1}{N} <\epsilon$.

I didn't understand how the $N$ is chosen here and used in the proof. (I understand the Archimedean Property but I'm confused of choosing $N$ in this aspect of the proof).

Answer 1

Let $\varepsilon>0$ be arbitrary. We want to show that there is an $N$ depending on $\varepsilon$ such that if $n\geq N$ then $\left| {\frac{1}{n} - 0} \right| <\varepsilon$. I claim that $N: = \left[ {1/\varepsilon } \right] + 1$ works. Indeed if $ n \ge N$ with this choice of $N$, then $$ \left| {\frac{1}{n} - 0} \right| = \frac{1}{n} \le \frac{1}{N} = \frac{1}{{\left[ {1/\varepsilon } \right] + 1}} < \frac{1}{{1/\varepsilon }} = \varepsilon . $$ In short, we make this choice of $N$ because it gives what we need. Another reasonable choice would be, for example, $N: = \left\lceil {1/\varepsilon } \right\rceil$. The only things that $N$ should satisfy are (i) $N$ has to be an integer (ii) $N$ has to be at least $1/\varepsilon$ (so that $1/N \leq \varepsilon$). $N: = \left[ {1/\varepsilon } \right] + 1$ is a convenient choice that satisfies both of these requirements.

Answer 2

Recall Archimedian priciple: given $x \in \mathbb{R}$, we can find $N \in \mathbb{N}$ such that $N > x$.

In particular, a remark is that if $x$ is nonnegative, we can always choose $N$ to be $[x] + 1$.

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For such thing, usually we work backward.

What do we want to find is $N(\epsilon)$ that would satisfy:

• $\forall \epsilon > 0 , \exists N \in \mathbb{N}, n \ge N \implies \left|\frac1n \right|<\epsilon. $

We can rewrite it as

• $\forall \epsilon > 0 , \exists N \in \mathbb{N}, n \ge N \implies n >\frac1\epsilon. $

Given $\epsilon > 0$, we want to choose $N \in \mathbb{N}$ such that $n\ge N$ then $n > \frac1\epsilon$. The Archimedian Principle of real number promises us that such $N \in \mathbb{N}$ exists such that $N > \frac1\epsilon$ (the $x$ in the Archimedian principle is $\frac1\epsilon$), hence if $n \ge N$, we can conclude that $n > \frac1{\epsilon}$.