B***y
发帖数: 83 | 1
If | b | > 1 then the series will be divergent.
If | b | < 1 then the series will be convergent.
The interesting case happens when |b| = 1, hmm... | B***y
发帖数: 83 | 2
Notice that it is a second order iteration equation, thus first write it as a
vector equation:
Y_(n+1) = A(n) * Y_(n) ,
where Y_n = (a(2n-1), a(2n))^t, where "t" means "transpose".
Easy to check that A(n) asymptotically behaves like (b 0
0 b )
Thus if |b| > 1, then the series Y_n will diverge, if |b|< 1 then the series
will converge. |
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