# Is 1/(n-2) convergent?

1. Jun 15, 2014

### kelvin490

Suppose there is a sequence xn=1/(n-2). We know we n tend to infinity the sequence tends to zero. But at n=2 it is equal to infinity. Is this sequence convergent?

There is also a theorem that all convergent sequence are bounded for every n. But the sequence above is not bounded at n=2.

From definition of convergent sequence it seems that only the case that n tends to infinity is concerned, it says nothing about whether it is convergent when n is finite but xn is not.

2. Jun 15, 2014

### HallsofIvy

I presume you are taking an introductory class in limits and are having trouble with the basic definitions because for years you have assumed things that are simply not true. "$\frac{1}{0}$ is NOT "infinity". It has NO value because "division by 0" is, literally, undefined.

The definition "says nothing about whether it is convergent when n is finite" because that makes no sense. We define convergence for a sequence only for the index going to infinity. Also it makes no sense to say that "$x_n$ is not finite" because the terms of a sequence are, by definition, real numbers and all real numbers are finite. It is NOT true that "at n= 2, 1/(n- 2) is equal to infinity". $x_2$ simply has NO value.

3. Jun 15, 2014

### kelvin490

Does it mean that for finite n, it is meaningless to say whether xn is convergent? Concept of convergence only applied for n tends to infinity?

In defining whether a sequence is bounded, n is for all values. Is xn bounded?

4. Jun 15, 2014

### micromass

Yes in the sense that if I'm only give finitely many $x_n$ then convergence is a meaningless thing. For example, if I'm only given $x_1,~x_2,~x_3,~x_4$, I cannot talk about convergence. I need to have $x_n$ close to infinity to discuss convergence.

Even better, if I leave out finitely many terms, that will not affect convergence. So if I'm given $x_1,~x_2,~x_3,...$ all the way to infinity, or if I'm given $x_4,~x_5,~x_6,...$ all the way to infinite, the concept of convergence will remain the same. If one sequence converges, so will the other and to the same value.

Of course, if I leave out infinitely many terms, then things can change. So the entire sequence $x_1,~x_2,~x_3,...$ might have a very different behavior than the sequence of odd terms $x_1,~x_3,~x_5,...$.

Boundedness behaves very similarly to convergence in the sense that if I leave out finitely many terms, that will not affect boundedness. But leaving out infinitely many terms might.

Although for a finite sequence it is meaningless to discuss convergence, we can discuss boundedness. But it will turn out a finite sequence is always bounded.

But wait, the example in your OP! We have $x_2 = \infty$ so it's not bounded! Well, first of all, like Halls remarked, $x_2$ is not infinity, it is simply undefined.
Furthermore, I would go further and say that the entire sequence is ill-defined. That is, if I look at the sequence $x_1,~x_2,~x_3,~x_4,~x_5,...$, then I'd say this sequence is not defined since one of the terms is undefined! So it is meaningless to discuss any properties of this sequence.

Of course, if we restrict the sequence to #x_3,~x_4,~x_5,...##, then everything is perfectly well-defined and we can talk about properties like convergence and boundedness.