Proof of divergence of (-1)^n sequence.

In summary, the proof for the divergence of the (-1) ^n sequence involves showing that there is no limit, by disproving the existence of a limit through the use of two conditions and the concept of epsilon. This is done by showing for any epsilon, there are always values in the sequence that do not satisfy the conditions, therefore proving that there is no number that can satisfy the definition of the limit.
  • #1
sabyakgp
4
0
Hello Friends,

I am at a loss to understand a proof concerning the proof of divergence of (-1) ^n sequence.
According to the book:
"To prove analytically that the sequence is convergent, it must satisfy both of the following conditions:

A: |-1-L| < epsilon
B: |+1 - L| < epsilon
"
(+1 and -1 are the only two values the sequence (-1)^n)

But, the book goes on, if we suppose epsilon = 1/2

"|-1-L| <1/2
which will not hold if L >0 since in that case |-1-L| = 1+L which is greater than 1/2
Likewise,
|+1-L| < 1/2 will not hold if L<0 since |+1-L| = 1 + |L| which is greater than 1/2.
Therefore since L can not be both negative and positive, there can be no limit to this sequence.
"
Though I understand the conditions, but fail to understand how it was proved that the limit does not exist. Could anyone please help me understand this?

Best Regards,
Sabya
 
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  • #2
sabyakgp said:
Hello Friends,

I am at a loss to understand a proof concerning the proof of divergence of (-1) ^n sequence.
According to the book:
"To prove analytically that the sequence is convergent, it must satisfy both of the following conditions:

A: |-1-L| < epsilon
B: |+1 - L| < epsilon
"
(+1 and -1 are the only two values the sequence (-1)^n)

But, the book goes on, if we suppose epsilon = 1/2

"|-1-L| <1/2
which will not hold if L >0 since in that case |-1-L| = 1+L which is greater than 1/2
Likewise,
|+1-L| < 1/2 will not hold if L<0 since |+1-L| = 1 + |L| which is greater than 1/2.
Therefore since L can not be both negative and positive, there can be no limit to this sequence.
"
Though I understand the conditions, but fail to understand how it was proved that the limit does not exist. Could anyone please help me understand this?

Best Regards,
Sabya
The presentation of the proof is a bit ugly I think, but the idea of disproving a limit is to show that for there is an epsilon for which any N you pick, there are some a_k > N that don't satisfy | a_k - L | < epsilon. Since 1 and -1 are our only values, picking any N we can just say our values of a_n after N are just 1 and -1.

Let's suppose that there is a limit of the sequence, call it L.
Pick epsilon to be 1/2. Suppose this limit is a nonnegative ( positive or zero ) number; then for any N, we have for some a_k > N, a_k = -1, so |a_k - L | = |-1-L | = |1 + L | since L is positive, we know that 1 + L >= 1, and | a_k - L | = | 1 + L | cannot possibly be less than 1/2. Therefore, our limit is not a positive number.
Now suppose the limit is negative. Remembering that our sequence is just alternating 1's and -1's, for any N, we can surely pick a_k > N so that a_k = 1. So, |a_k - L | =
| a_k + |L| | = | 1 + |L| | = 1 + |L| ( remember that L is negative ). Picking epsilon to be 1/2 again shows us that | a_k - L | = 1 + |L| < 1/2 can never happen.

Therefore, if L is a non-negative number there is an epsilon so that for any N , there is an a_k > N so that |a_k - L | > epsilon. The same happens if L is negative. Therefore, there is no number L that can satisfy the definition of the limit of this sequence
 

What is the (-1)^n sequence?

The (-1)^n sequence is a mathematical sequence where each term alternates between positive and negative values. The value of n determines whether the term is positive or negative, with even values resulting in positive terms and odd values resulting in negative terms.

What is proof of divergence?

Proof of divergence is a mathematical concept used to show that a sequence does not have a finite limit. In other words, the terms in the sequence do not approach a specific value as n approaches infinity.

How do you prove divergence of the (-1)^n sequence?

To prove divergence of the (-1)^n sequence, you can use the limit comparison test or the direct comparison test. Both methods involve comparing the given sequence to a known divergent sequence, such as the harmonic series.

Why is proving divergence important?

Proving divergence is important because it allows us to determine whether a given sequence has a finite limit or not. This information is useful in various mathematical applications, such as determining the convergence of series or evaluating the behavior of a function at certain points.

What are some real-world applications of the (-1)^n sequence?

The (-1)^n sequence can be used to model alternating current in electrical circuits, where the direction of the current changes periodically. It can also be used in economics to represent alternating profits and losses in a business. Additionally, the sequence can be used to study the convergence and divergence of various series in mathematics and physics.

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