Determining of a sequence is convergent or divergence

In summary, the problem is to determine the convergence or divergence of the sequence x_{n} := (-1)^{n}n/(n+1), and the student is having trouble using the squeeze theorem to solve it. They have already shown that x_{n} := (-1)^{n}/(n+1) converges to zero, but the additional "n" in the numerator is causing difficulties. They have tried writing out several terms in the sequence, but are still unable to determine its behavior. They are seeking guidance on how to use a theorem to prove that the sequence is divergent.
  • #1
nlsherrill
323
1

Homework Statement


For x[itex]_{n}[/itex] given by the following formula, establish either the convergence or divergence of the sequence [itex]X = (x_{n})[/itex]

[itex]x_{n} := (-1)^{n}n/(n+1)[/itex]

Homework Equations


The Attempt at a Solution


This is for my real analysis class. I tried to use the squeeze theorem, but didn't get anywhere with it. I know [itex](-1)^{n}[/itex] is divergent, and n are divergent sequences but I haven't been able to use many theorems in the section because most of them are assuming something about a convergent sequence(and I need to show if it is or isn't)

Any ideas?

Thanks for lookingedit:

*Note: I have already shown that [itex]x_{n} := (-1)^{n}/(n+1)[/itex] converges to zero using the squeeze theorem, but the extra "n" in the numerator is messing me up with this one..
 
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  • #2
nlsherrill said:

Homework Statement


For x[itex]_{n}[/itex] given by the following formula, establish either the convergence or divergence of the sequence [itex]X = (x_{n})[/itex]

[itex]x_{n} := (-1)^{n}n/(n+1)[/itex]


Homework Equations





The Attempt at a Solution


This is for my real analysis class. I tried to use the squeeze theorem, but didn't get anywhere with it. I know [itex](-1)^{n}[/itex] is divergent, and n are divergent sequences but I haven't been able to use many theorems in the section because most of them are assuming something about a convergent sequence(and I need to show if it is or isn't)

Any ideas?

Thanks for looking


edit:

*Note: I have already shown that [itex]x_{n} := (-1)^{n}/(n+1)[/itex] converges to zero using the squeeze theorem, but the extra "n" in the numerator is messing me up with this one..
Write out about a dozen terms in the sequence. That should give you a better idea about what this one is doing.
 
  • #3
Mark44 said:
Write out about a dozen terms in the sequence. That should give you a better idea about what this one is doing.

well sure I can do that and tell that it is heading towards -1 and 1, i.e. its divergent, but I was assuming the problem was asking some kind of use of a theorem to prove its divergent
 
  • #4
You could use the definition of a divergent sequence (the negation of the definition of a convergent sequence).
 

1. How do you determine if a sequence is convergent or divergent?

To determine if a sequence is convergent or divergent, you need to take the limit of the sequence as it approaches infinity. If the limit exists and is a finite number, the sequence is convergent. If the limit does not exist or is infinite, the sequence is divergent.

2. What is the difference between a convergent and divergent sequence?

A convergent sequence is one where the terms of the sequence approach a specific, finite number as the number of terms increases. A divergent sequence, on the other hand, does not approach a specific number as the number of terms increases. It either approaches infinity or does not have a limit at all.

3. Can a sequence be both convergent and divergent?

No, a sequence can only be either convergent or divergent. It cannot be both at the same time. If the limit of a sequence exists and is a finite number, it is convergent. If the limit does not exist or is infinite, it is divergent.

4. How do you prove that a sequence is convergent or divergent?

To prove that a sequence is convergent, you can use the definition of convergence and show that the limit of the sequence as it approaches infinity exists and is a finite number. To prove that a sequence is divergent, you can use the definition of divergence and show that the limit does not exist or is infinite.

5. Are there any special cases where a sequence can be both convergent and divergent?

No, there are no special cases where a sequence can be both convergent and divergent. A sequence can only be either convergent or divergent, as determined by the limit of the sequence as it approaches infinity.

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