- #1
phospho
- 251
- 0
The definition of convergence is given by : ## \forall \epsilon > 0, \exists N \in \mathbb{R} ## such that ## |x_n - l | < \epsilon ## ## \forall n \in \mathbb{N} ## with ## n > N ##
negate this statement and prove that the sequence ## x_n = (-1)^nn ## is divergent using only the negation of the definition of convergence
ok here's my attempt of the negation of the statement:
## \exists \epsilon > 0 \forall N \in \mathbb{R} ## s.t. ## \exists n \in \mathbb{N} n > N ## and ## |x_n - l | \geq \epsilon ##
proof that ## x_n = (-1)^nn ## does not converge:
since the negation of the definition is "## \forall N ##" let N = 1, and find ## n_1 > 1 ## then let ## N = n_1 ## and find ## n_2 > N=n_1 ## etc leaving us with a sequence of indicies ## n_1 < n_2 < n_3.. ##
so let's prove that it does not converge to 1:
choose ## epsilon = 1/2 ## then for ## N>0 ## let n be the odd integers i.e. ## n_1 = 1 < n_2 = 3 < n_3 = 5 ... ## then n > N hence ## | (-1)^1(1) - 1 | >= 1/2 ## which is true hence ## x_n## does not converge to 1
take ## l > 1 ## and choose ##\epsilon ## s.t. ## \epsilon < l - 1 ## again, choosing ## N > 0 ## and n being all the odd integers greater than 0 we have n > N and ## |(-1)(1) - l | \geq \epsilon## which is true
take ## l < 1## and choose n to be all the odd integers above 0 then n > N and ## |-1 - l | >= \epsilon ## which is true therefore ## x_n ## does not converge
I think this works, but truth be told, I don't understand the negation of this statement. I mean, I understand HOW I got the negation, but I don't know what it means. The choosing N > 0 is from my lecture notes and I don't understand why that works, and why we can just choose N > 0 if the definition states for all N
any help please
negate this statement and prove that the sequence ## x_n = (-1)^nn ## is divergent using only the negation of the definition of convergence
ok here's my attempt of the negation of the statement:
## \exists \epsilon > 0 \forall N \in \mathbb{R} ## s.t. ## \exists n \in \mathbb{N} n > N ## and ## |x_n - l | \geq \epsilon ##
proof that ## x_n = (-1)^nn ## does not converge:
since the negation of the definition is "## \forall N ##" let N = 1, and find ## n_1 > 1 ## then let ## N = n_1 ## and find ## n_2 > N=n_1 ## etc leaving us with a sequence of indicies ## n_1 < n_2 < n_3.. ##
so let's prove that it does not converge to 1:
choose ## epsilon = 1/2 ## then for ## N>0 ## let n be the odd integers i.e. ## n_1 = 1 < n_2 = 3 < n_3 = 5 ... ## then n > N hence ## | (-1)^1(1) - 1 | >= 1/2 ## which is true hence ## x_n## does not converge to 1
take ## l > 1 ## and choose ##\epsilon ## s.t. ## \epsilon < l - 1 ## again, choosing ## N > 0 ## and n being all the odd integers greater than 0 we have n > N and ## |(-1)(1) - l | \geq \epsilon## which is true
take ## l < 1## and choose n to be all the odd integers above 0 then n > N and ## |-1 - l | >= \epsilon ## which is true therefore ## x_n ## does not converge
I think this works, but truth be told, I don't understand the negation of this statement. I mean, I understand HOW I got the negation, but I don't know what it means. The choosing N > 0 is from my lecture notes and I don't understand why that works, and why we can just choose N > 0 if the definition states for all N
any help please