Is the Sequence xn=1/(n-2) Convergent for All Values of n?

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In summary, the conversation discusses the concept of convergence and boundedness for sequences. It clarifies that convergence is only defined for the limit as n tends to infinity, and that leaving out finitely many terms does not affect convergence or boundedness. The example given in the original post is not well-defined since one of the terms is undefined, making it meaningless to discuss any properties of the sequence.
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kelvin490
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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.
 
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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. "[itex]\frac{1}{0}[/itex] 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 "[itex]x_n[/itex] 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". [itex]x_2[/itex] simply has NO value.
 
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HallsofIvy said:
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. "[itex]\frac{1}{0}[/itex] 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 "[itex]x_n[/itex] 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". [itex]x_2[/itex] simply has NO value.

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?
 
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kelvin490 said:
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?

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,...##.

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

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.
 
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Therefore, we cannot determine whether the sequence xn=1/(n-2) is convergent or not based on the given information. In order to determine its convergence, we would need to know the behavior of the sequence for all values of n, not just when n tends to infinity.
 

Related to Is the Sequence xn=1/(n-2) Convergent for All Values of n?

What is the definition of convergence in mathematics?

In mathematics, convergence refers to the property of a sequence or series of numbers that approaches a specific limit as the number of terms increases. In other words, as the value of the terms gets closer and closer to a certain number, the sequence or series is said to converge.

What is the significance of the number "n-2" in determining convergence?

The number "n-2" is significant because it represents the number of terms in the sequence or series that are being added or subtracted. In order for a sequence or series to converge, the number of terms must approach infinity, which means that the value of "n-2" must be very large. This helps to determine whether or not the sequence or series will converge.

How do you determine if 1/(n-2) is a convergent sequence?

To determine if the sequence 1/(n-2) is convergent, you can use the limit comparison test. This test compares the given sequence to a known convergent or divergent series to determine its convergence. If the limit of the given sequence is equal to the limit of the known series, then the given sequence is convergent. If the limits are not equal, then the sequence is divergent.

What are the possible values of "n" that would make 1/(n-2) a convergent series?

The series 1/(n-2) will only converge if the value of "n" is greater than or equal to 3. This is because as "n" approaches 2, the denominator becomes 0, which is undefined. Therefore, the series will only converge if "n" is greater than 2.

What are the implications of 1/(n-2) being a convergent series?

If 1/(n-2) is a convergent series, it means that the terms in the series are approaching a specific limit as the number of terms increases. This can have important implications in various mathematical applications, such as calculating the sum of infinite series or determining the convergence of integrals. It also indicates that the series has a finite sum, which can be useful in solving mathematical problems.

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