f**n
发帖数: 401 | 1 Define A_n=\sum_{i=1}^n a_i, B_n=\sum_{i=1}^n b_i in which a_i and b_i are
both 无穷小量 or equivalently:
\lim_{n->\infty}a_i=\lim_{n->\infty}b_i=0
Moreover for the same i, a_i and b_i are 等价无穷小 or equivalently:
\lim_{n->\infty}a_i/b_i=1, 1<=i<=n
Now we want to prove:
\lim_{n->\infty}A_n=\lim_{n->\infty}B_n if either of the limits exists.
The above equation seems intuitively true...to me. But
it seems that the proof requires the exchange of \lim and \sum.
I am not sure the requirements for this exch | I***e
发帖数: 1136 | 2
I don't think the conclusion is right. Take a1=1 and ai=0 for i>1. Take bi=0
for all i.
THen An = 1 and Bn=0 for all n.
How can lim An = lim Bn ?
I guess you meant An= sum_{i=n}^infty ai. If so, you have to also have some
extra conditions like ai, bi are both positive sequences. Then you can use the
discrete version of BCT ( Bounded convergence theorem )
Hope this helps.
icare
【在 f**n 的大作中提到】
: Define A_n=\sum_{i=1}^n a_i, B_n=\sum_{i=1}^n b_i in which a_i and b_i are
: both 无穷小量 or equivalently:
: \lim_{n->\infty}a_i=\lim_{n->\infty}b_i=0
: Moreover for the same i, a_i and b_i are 等价无穷小 or equivalently:
: \lim_{n->\infty}a_i/b_i=1, 1<=i<=n
: Now we want to prove:
: \lim_{n->\infty}A_n=\lim_{n->\infty}B_n if either of the limits exists.
: The above equation seems intuitively true...to me. But
: it seems that the proof requires the exchange of \lim and \sum.
: I am not sure the requirements for this exch
| f**n
发帖数: 401 | 3 Yes. Sorry about that.
Let me try to make it clearer:
Actually I am trying to emulate an integral by a summation:
You can imagine there is a interval(x axis) and I divide the interval into
n smaller intervals. After this decomposition, we index all intervals, let's
say, interval 1,2,...,n. For interval i, a_i and b_i are defined.
So if n->\infty, both series {a_i} and {b_i} have to evolve. |
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