and define

Prove

Is it as simple as

Therefore,

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- Jun 18th 2011, 07:18 PMdwsmith{a_n}\to A and define {b_n}
and define

Prove

Is it as simple as

Therefore, - Jun 18th 2011, 08:58 PMTheEmptySetRe: {a_n}\to A and define {b_n}
- Jun 19th 2011, 07:21 AMdwsmithRe: {a_n}\to A and define {b_n}
I don't know why you picked and I don't know where to go from the .

Can you elaborate further. - Jun 19th 2011, 07:48 AMPlatoRe: {a_n}\to A and define {b_n}
- Jun 19th 2011, 08:46 AMDevenoRe: {a_n}\to A and define {b_n}
you can rewrite this difference as the sum of two differences, and use a version of the triagle inequality.....

the "idea" behind this is that any convergent series is Cauchy, so for large enough n, two successive terms are as close as we like (this means something in terms of epsilon), so the "average" of the two successive terms (for large enough n) is going to be close to the limit as well. - Jun 19th 2011, 06:13 PMdwsmithRe: {a_n}\to A and define {b_n}
- Jun 19th 2011, 07:12 PMDevenoRe: {a_n}\to A and define {b_n}
since {an}-->A, for any ε > 0, there IS some N with |an - A| < ε for all n > N. so 1/2 of that is less than ε/2.

if n > N, surely n+1 > N as well, thus |a(n+1) - A| < ε, too. there is no need for a "secondary" epsilon: