Show that (-1)^n * (n/(n+1)) is divergent

  • Thread starter PhyzX
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Of course you can use subsequences. The whole point of using subsequences is that you can use them to prove a sequence doesn't converge. What you can't use is a theorem that says "if a sequence is Cauchy, it converges". That's because showing that a sequence is Cauchy is essentially the same as showing it converges, if the space you're working in is complete. But showing that a sequence has *no* limit is not the same thing as showing it has a limit.
  • #1

PhyzX

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Homework Statement


Using a proof by cases, show that (-1)^n * (n/(n+1)) is divergent


Homework Equations



A sequence a_n is said to be convergent if
For every real number ε > 0, there exists a natural number N such that for all n > N, |a_n − L| < ε.

The Attempt at a Solution


Tried to make n the next odd integer (2k+1) but struggled to go any further.
 
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  • #2
Your expression is unclear (to me) ... is it supposed to be a summation? Is the n/(n+1) a factor or part of the exponent?

At a guess you mean the sum -1/2 + 2/3 - 3/4 + 4/5 ... starting with n=1.

Also I don't understand what you mean by "cases" ... what would the cases be?
 
  • #3
UltrafastPED said:
Your expression is unclear (to me) ... is it supposed to be a summation? Is the n/(n+1) a factor or part of the exponent?

At a guess you mean the sum -1/2 + 2/3 - 3/4 + 4/5 ... starting with n=1.

Also I don't understand what you mean by "cases" ... what would the cases be?

Sorry if I wasn't clear. The expression is a sequence starting from 1 through the Natural numbers. n/(n+1) is a factor not part of the exponent.
The cases are
L<0
L=0
L>0

Where L is the limit of the sequence.
 
  • #4
A sequence is a list of numbers {1,2,3,...}; it may have a limit point, but not a limit.

An infinite sum is defined to be the limit point of its sequence of partial sums; the infinite summation does not have a limit - it either has a value or is divergent.

So I don't understand what L is supposed to represent - though I do understand the case structure.

Please write out a few of the terms ... was my guess correct?
 
  • #5
UltrafastPED said:
A sequence is a list of numbers {1,2,3,...}; it may have a limit point, but not a limit.

An infinite sum is defined to be the limit point of its sequence of partial sums; the infinite summation does not have a limit - it either has a value or is divergent.

So I don't understand what L is supposed to represent - though I do understand the case structure.

Please write out a few of the terms ... was my guess correct?

Your guess is correct but it is not a sum, just a sequence. I had a look around and found a similar question albeit slightly simpler: https://www.physicsforums.com/showthread.php?t=491938
I'm struggling to emulate the techniques used in that thread.
 
  • #6
The sequence is {-1/2, +2/3, -3/4, +4/5, ...} with an assumed limit point L.

Proof that this sequence diverges by the method of contradiction in three cases: limit point L assumed to by +,0,-.

Is that the problem?
 
  • #7
UltrafastPED said:
The sequence is {-1/2, +2/3, -3/4, +4/5, ...} with an assumed limit point L.

Proof that this sequence diverges by the method of contradiction in three cases: limit point L assumed to by +,0,-.

Is that the problem?

Yes that is the problem, sorry it took so long to get across.
 
  • #8
The key to mathematics is clarity, with brevity a close second ... hence the importance of definitions.

I looked at the referenced thread ... I suggest you study a clearer example. Try this one:
http://www.westmont.edu/~howell/courses/ma-108/illustrations/text-solutions/page189-6.pdf
 
  • #9
UltrafastPED said:
The key to mathematics is clarity, with brevity a close second ... hence the importance of definitions.

I looked at the referenced thread ... I suggest you study a clearer example. Try this one:
http://www.westmont.edu/~howell/courses/ma-108/illustrations/text-solutions/page189-6.pdf

Thanks for the link, however I am required to use the contradiction in each case method.
 
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  • #10
UltrafastPED said:
Your expression is unclear (to me) ... is it supposed to be a summation? Is the n/(n+1) a factor or part of the exponent?
The sequence as described in the OP is clear to me.
UltrafastPED said:
At a guess you mean the sum -1/2 + 2/3 - 3/4 + 4/5 ... starting with n=1.
There was nothing in the OP that would lead someone to think that this was a problem about a series.
UltrafastPED said:
Also I don't understand what you mean by "cases" ... what would the cases be?

UltrafastPED said:
A sequence is a list of numbers {1,2,3,...}; it may have a limit point, but not a limit.
?
It is quite common to talk about the limit of a sequence. For example, with the sequence {an}, where an = ##\frac{n}{n + 1}, \lim_{n \to \infty}a_n = 1##.
UltrafastPED said:
An infinite sum is defined to be the limit point of its sequence of partial sums; the infinite summation does not have a limit - it either has a value or is divergent.

So I don't understand what L is supposed to represent - though I do understand the case structure.

Please write out a few of the terms ... was my guess correct?
 
  • #11
PhyzX, if you haven't already done so, look at the first five or six terms in your sequence. The terms jump around, due to the (-1)n factor.
 
  • #12
Mark44 said:
PhyzX, if you haven't already done so, look at the first five or six terms in your sequence. The terms jump around, due to the (-1)n factor.

Hey, yeah I've noticed that. The sequence converges to both 1 and -1 if I'm not mistaken? However I am told to use the method of finding suitable epsilon and N in each case to produce counter examples. So I can't use subsequences and such to show that it converges twice.
However I think I've got it now.

After I substitute a value for n, say the first even integer greater than N, (n=2k), I'm left with something like | (2k)/(2k+1) - L | and since k is greater than 1, I can manipulate the terms to show that it is bigger than say ε=1/4
 
  • #13
PhyzX said:
Hey, yeah I've noticed that. The sequence converges to both 1 and -1 if I'm not mistaken?
No, that's not right. If a sequence converges, it can converge to only one value. It can't converge to both 1 and -1. Your sequence has a subsequence that converges to 1 and another that converges to -1, but the sequence itself doesn't converge.
PhyzX said:
However I am told to use the method of finding suitable epsilon and N in each case to produce counter examples. So I can't use subsequences and such to show that it converges twice.
However I think I've got it now.

After I substitute a value for n, say the first even integer greater than N, (n=2k), I'm left with something like | (2k)/(2k+1) - L | and since k is greater than 1, I can manipulate the terms to show that it is bigger than say ε=1/4

Keep in mind that when N is large, the terms in your sequence are bouncing around between numbers that are close to 1 and -1. When one term in your sequence is close to +1, the next term is going to be almost 2 units away, near -1. Similarly, when one term in the sequence is near -1, the next term will be again almost 2 units away, near +1.
 

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