Does Monotone Convergence imply Convergence Subsequence?

annoymage
Messages
360
Reaction score
0

Homework Statement



Results

i) if (a_n) tends to L as n tends to infinity, then a_{n_r} tends to L as r tend to infinity

ii)if (a_n) tends to infinity as n tends to infinity, then a_{n_r} tends to infinity as r tend to infinity

using this result prove that

if (a_n) is an increasing sequence, prove that the converse of i) is true
Suppose (a_n) is divergent, then by ii), all the subsequences must be divergent, so, cant.

therefore (a_n) must be convergent, means (a_n) tends to M for some M,

apply i), then means M=L

but how come i didn't use the fact (a_n) is monotone, must be something wrong somewhere, help T_T
 
Physics news on Phys.org
"Tends to infinity" is not the only way a sequence can diverge. Oscillating sequences are also divergent.
 
aaaaaaaaaaaarghh, yes yes, thank you ^^

hmm, now i have to prove that <br /> (a_n)<br /> is convergent,

i suspect i should prove that <br /> (a_n)<br /> is bounded then, i know <br /> (a_n)<br /> is monotone then, <br /> (a_n)<br /> must be converging right? then continue like i was doing above right?
 
Last edited:
That sounds correct.
 
thank you very much
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top