MQ80 --- Sequence

Set a(n)=(-1)^n.
If lim(n->inf) a(n) exists, then lim(n->inf) [a(2n)-a(2n-1)]=0.
Now, lim(n->inf) [a(2n)-a(2n-1)] = 2≠ 0,
so that lim(n->inf) a(n) does not exist, ie. { (-1)^n} diverges.

2013-04-10 14:52:32 補充：
If lim(n->∞) a(n) = lim(n->∞) (-1)^n = L (exists), then
there exists M in N such that for all n>M, | a(n)-L | < 1/2 (taking ε = 1/2),
so that, |a(n)-a(n+1)| <= | a(n) - L | + | a(n+1) - L| < 1/2 + 1/2 = 1, for n>M
thus | a(n)- a(n+1) | = 1 < 1 (impossible)
Hence, lim(n->∞) a(n) does not exist.

2013-04-10 20:49:18 補充：
If lim(n->∞) a(n) = L (exists), then
there exists M in N such that for all n>M, | a(n)-L | < 1/2 (taking ε = 1/2),
so that, |a(n)-a(n+1)| <= | a(n) - L | + | a(n+1) - L| < 1/2 + 1/2 = 1, for n>M
thus | a(n)- a(n+1) | = 1 < 1 (impossible)
Hence, lim(n->∞) a(n) does not exist.

2013-04-11 17:15:28 補充：
If lim(n->∞) a(n) = L (exists), then
there exists M in N such that for all n>M, | a(n)-L | < 1/2 (taking ε = 1/2),
so that, |a(n)-a(n+1)| <= | a(n) - L | + | a(n+1) - L| < 1/2 + 1/2 = 1, for n>M
thus | a(n)- a(n+1) | = 2 < 1 (impossible)
Hence, lim(n->∞) a(n) does not exist.

Isn't |a(n) - a(n + 1)| = 2 ?