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

• PhyzX
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.

## 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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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?

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.

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?

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.

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?

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.

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

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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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?

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.

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

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.