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

[sabyakgp]

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

 

[wisvuze]

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