suppose limit a_n = A
prove that limit (a_n+1 - a_n) = 0
many thanks
2006-10-12 20:04:04 · 4 answers · asked by superfly 1 in Science & Mathematics ➔ Mathematics
Let e > 0. There exists a natural number N such that
|a_n - A| < e/2 and |a_n+1 - A| < e/2. It follows that
|a_n+1 - a_n| =
|a_n+1 - A + A - a_n| <=
|a_n+1 - A| + |a_n - A| <
e/2 + e/2 < e.
By the definition of a limit of a sequence, a_n+1 - a_n ->0 as n -> infinity.
2006-10-12 20:08:32 · answer #1 · answered by James L 5 · 1⤊ 0⤋
lim a_n = A
as A is independent of n so
lim A_n+1 is A
so lim (a_n+1 - a_n)
= lim a_n+1 - lim a_n = A - A =0
2006-10-13 04:48:17 · answer #2 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
because the limit of the sum is the sum of the limits. yuoi dont rearrange the terms
2006-10-13 05:41:00 · answer #3 · answered by gjmb1960 7 · 0⤊ 0⤋
o_*=(758348477"-) your welcome.
2006-10-13 03:07:39 · answer #4 · answered by triniqueen40 4 · 0⤊ 0⤋