How to prove $\frac{1}{n}$ converges to 0

Question

How do I prove $\frac{1}{n}$ converges to 0?

my attempt:

for $\epsilon >0$, let $N \geq \frac{1}{\epsilon}$

if $n > N \geq \frac{1}{\epsilon}$ then $\epsilon > \frac{1}{n} = \left| \frac{1}{n} \right| = \left| \frac{1}{n} - 0\right| $

so $\frac{1}{n}\to 0$

But how do we prove it geometrically?

And how do we prove $\frac{1}{n}$ does not converge to any non zero number?

For example, if we choose $l=\frac{1}{10}$ then how do we choose a number $\epsilon$ such that $x_n=\frac{1}{n} \notin(l-\epsilon, l+\epsilon)$?

Why do you want a "geometric" proof? It's not the 1600s anymore. — hmakholm left over Monica, Aug 10 '18 at 10:51
You can choose $\epsilon=|l-0|/4$. Then for pick $N\geq1/\epsilon$. It follows that for all $n>N$ $0<1/n<\epsilon=|l|/4$ is outside $(l-\epsilon,l+\epsilon)$. — , Aug 10 '18 at 10:56
@henning makholm... Just doubt btw I m still still studying 1600 maths — Inverse Problem, Aug 10 '18 at 10:57
To produce a geometric proof you only need to pick [[any geometric construction of the reciprocal]] and repeat the same argument that you did on choosing a segment of integer length $N>1/\epsilon$. — , Aug 10 '18 at 11:04
@fatherBrown You could potentially adapt that construction to this triangle and show that as the segment $AD=n$ tends to $+\infty$, the line $GB$ becomes coincident with $AD$ and the segment $AG=\frac1n$ tends to $0$. https://imgur.com/ZK1KmLk — Jam, Aug 10 '18 at 11:34
Or equivalently, given any $\varepsilon$ on the segment $AE$, it's possible to make $AG<A\varepsilon$ by making $AD$ sufficiently large. — Jam, Aug 10 '18 at 11:36

score 3 · Accepted Answer · answered Aug 10 '18 at 10:56

3

One of the basic theorems about limits is that if a limit exists then it is unique. If your sequence converges to $0$ then it can't converge to anything else. Try to prove that theorem, it's a good exercise. And it will answer your question.

answered Aug 10 '18 at 10:56

Mark

43,582

score 2 · Answer 2 · answered Aug 10 '18 at 11:13

2

By contradiction suppose $\frac1n\to L\neq 0$ and apply the definition.

answered Aug 10 '18 at 11:13

user

162,563

score 1 · Answer 3 · answered Aug 10 '18 at 13:40

A mathematical proof is always based on axioms: you prove that a less obvious fact is true based on the assumption that more obvious facts are true. This is why sometimes when we try to prove very obvious things, we get this weird gut feeling like something must be wrong, like our reasoning must be bad or circular. It's because since the statement we're trying to prove is so obvious, the axioms we used are probably not all that much more obvious, and if we're not careful they might even be less obvious. So let's take a look at exactly what axioms you're implicitly using here.

You argument is that for $n>\frac 1\epsilon$, we will have $\frac 1 n < \epsilon$. Since $\frac1n$ is obviously not negative, this means $|\frac1n-0|<\epsilon$. The exact assumptions you're using here are that:

$\frac1x$ is a decreasing function.
$\frac1x$ is injective and is its own inverse function.
$\frac1x$ maps positive reals to positive reals. (This assumption is important: without it, the conclusion that $\frac1n<\epsilon$ tells you nothing about convergence to zero since it could just mean that $\frac1n$ is negative ten billion).

It is in fact true that any function satisfying these conditions must converge to zero (an interesting little result on its own), so your proof is fine. Are you happy with these axioms? After all, you could have saved yourself some trouble by just assuming without proof that $\frac1n\to0$, so do the three assumptions above seem more reasonable to you than just assuming that? Then your proof is fine.

How to prove $\frac{1}{n}$ converges to 0

my attempt:

3 Answers3