Re: find the limit of this quotient
- From: magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin)
- Date: Tue, 4 Mar 2008 15:23:16 +0000 (UTC)
In article <ff6fc98f-5a1a-455b-8b89-e710fbfb9b86@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<taxxon1987@xxxxxxxxx> wrote:
[...]
Question: What are the possible limits which can take the quotient
which consists of the sequence
a_(n+1)/a_n when n tends to infinity?
[...]
this is interesting. But just wondering if you can generalize the
idea. As in can you always prove that if the sequence converges to
zero (lim a_n=0), there will be a subsequence a_n_k such that a_n_(k
+1)/a_n_k=0?
Assuming none of the terms are zero, presumably, so that the quotient
makes sense.
Yes. Since the sequence convergest to 0, there exists N_1 such that
|a_n|<1/2 for all n>= N_1.
Likewise, there exists an N_2 such that for all n>= N_2,
|a_n| < |a_{N_1}|/2^1. Note that N_2 > N_1.
And there exists N_3 such that for all n>= N_3 we will have
|a_n| < | a_{N_2}|/2^2.
Assuming we have defined N_k so that for i=2,...,k, if n>=N_i, then
|a_n| < | a_{N_{i-1}}|/2^{i-1}, we define N_{k+1} so that if
n>=N_{k+1}, then |a_n| < | a_{N_k}|/2^k.
Thus we obtain a subsequence a_{N_1}, a_{N_2}, ..., a_{N_k}, ... of
the original sequence. And
|a_{N_{k+1}}|/|a_{N_k}| <= |a_{N_k}/2^{k}|/|a_{N_k}| = 1/2^k
Therefore, the limit as k->oo of a_{N_{k+1}}/a_{N_k} is equal to 0.
--
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what I accept as reality."
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Arturo Magidin
magidin-at-member-ams-org
.
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