# Real analysis, sequence of sequences convergence proof

Dick
Homework Helper
i dont get it since if im not mistaken then there is lots of more sequences that converges to zero, and the sequence in L that converges to it would probably not look like post #1 sequence at all. i might have some holes in my understanding of this so don't hesitate by giving me details that you might think i already know
Hmm. Suggest a sequence from ##\ell## that might converge to (1,1/2,1/4,1/8,1/16,...).

Hmm. Suggest a sequence from ##\ell## that might converge to (1,1/2,1/4,1/8,1/16,...).
well if ##u_k = (1/2^0, 1/2^1 , 1/2^2 , 1/2^3, 1/2^4 ... 1/2^k...0,0,0, ... )##
then the sequence ##u_k## from k=0 to infinity will converge to that ?

Dick
Homework Helper
well if ##u_k = (1/2^0, 1/2^1 , 1/2^2 , 1/2^3, 1/2^4 ... 1/2^k...0,0,0, ... )##
then the sequence ##u_k## from k=0 to infinity will converge to that ?
Perfect. You still don't see what to do with a general sequence that converges to zero? Take ##(a_1,a_2,a_3,...)##.

Perfect. You still don't see what to do with a general sequence that converges to zero? Take ##(a_1,a_2,a_3,...)##.
i still dont see it. what properties does this sequence (a1,a2,a3....) have ? is it like a special kind of general structure of all these zero convergent sequences. Because i dont see anything general about those two concrete examples you gave me or how they are relevant. I'm sure there are more zeorconvergen sequences that is not on the form 1/k^n somthing , . i would think a zeroconvergence sequence could be anything, only thing that's special about it is that it gets closed to zero

Dick
Homework Helper
i still dont see it. what properties does this sequence (a1,a2,a3....) have ? is it like a special kind of general structure of all these zero convergent sequences. Because i dont see anything general about those two concrete examples you gave me or how they are relevant. I'm sure there are more zeorconvergen sequences that is not on the form 1/k^n somthing , . i would think a zeroconvergence sequence could be anything, only thing that's special about it is that it gets closed to zero
Hmm. Ok, ##a=(a_1,a_2,a_3,...)##, a sequence that converges to 0. How about if you define ##u_k=(a_1,a_2,...,a_k,0,0,0,...)##? What is the limit of the sequence ##u_k##?

Hmm. Ok, ##a=(a_1,a_2,a_3,...)##, a sequence that converges to 0. How about if you define ##u_k=(a_1,a_2,...,a_k,0,0,0,...)##? What is the limit of the sequence ##u_k##?
it will converges to ##a##, i think i found some kind of proof now.

let ##a = (a_1,a_2,a_3,a_k..)## be a zeroconvergent sequence.
then for all ##n,m > N , d(a_n, a_m)<\epsilon##, for some ##N## then
let ##n## be ##n+1## and ##m## be ##n+a## s.t ##d(a_{n+1}, a_{n+a})<\epsilon.##

then the sequence of sequences ##L = (a_1,a_2,...,a_k,0,0,0,...)## for k=1 to infinity in ##\ell## would converge to ##a##
because ##d(L_n , L_{n+a}) = sup|L_n - L_{n+a}|## which all the elements up to and including ##L_n## turns zero and you end up with the supremum of the difference, ##L_{n+1}## must be the supremum, since its a decreasing sequence.
since ##L_{n+1}## is the ##a_{n+1}## of the sequence ##a##, and ## L_{n+a} = a_{n+a} ## then
##d(L_{n+1} , L_{n+a}) < \epsilon##. for all ##n>N##

Dick
Homework Helper
it will converges to ##a##, i think i found some kind of proof now.

let ##a = (a_1,a_2,a_3,a_k..)## be a zeroconvergent sequence.
then for all ##n,m > N , d(a_n, a_m)<\epsilon##, for some ##N## then
let ##n## be ##n+1## and ##m## be ##n+a## s.t ##d(a_{n+1}, a_{n+a})<\epsilon.##

then the sequence of sequences ##L = (a_1,a_2,...,a_k,0,0,0,...)## for k=1 to infinity in ##\ell## would converge to ##a##
because ##d(L_n , L_{n+a}) = sup|L_n - L_{n+a}|## which all the elements up to and including ##L_n## turns zero and you end up with the supremum of the difference, ##L_{n+1}## must be the supremum, since its a decreasing sequence.
since ##L_{n+1}## is the ##a_{n+1}## of the sequence ##a##, and ## L_{n+a} = a_{n+a} ## then
##d(L_{n+1} , L_{n+a}) < \epsilon##. for all ##n>N##
It doesn't say that the sequence ##a## is decreasing, it just says that it converges to zero. I think you've got the right idea but your notation is pretty confusing. If you want to set ##L_k = (a_1,a_2,...,a_k,0,0,0,...)## then you want to prove that the sequence ##(L_1,L_2,L_3,...)## converges to ##a##. So for all ##\epsilon \gt 0## there exists an ##N## such that for all ##n \gt N## that ##d(L_n,a) \lt \epsilon##. Can you try to state a proof of that a bit more clearly?

It doesn't say that the sequence ##a## is decreasing, it just says that it converges to zero. I think you've got the right idea but your notation is pretty confusing. If you want to set ##L_k = (a_1,a_2,...,a_k,0,0,0,...)## then you want to prove that the sequence ##(L_1,L_2,L_3,...)## converges to ##a##. So for all ##\epsilon \gt 0## there exists an ##N## such that for all ##n \gt N## that ##d(L_n,a) \lt \epsilon##. Can you try to state a proof of that a bit more clearly?

let ##a = (a_1,a_2,a_3,a_k..)## be a zero-convergent sequence.
then the sequence of sequences ##L_n = (a_1,a_2,...,a_n,0,0,0,...)## for n=1 to infinity in ##\ell## would converge to ##a##
because ##d(L_n , a) = sup|L_n - a|##, all the elements up to and including ##a_n## turns to zero, therfor ##d(L_n , a) = sup|(a_{n+1}, a_{n+2}...)|##
since for all ##n>N, d(a_n,0) < \epsilon## for some N then
##d(L_n , a) = sup|(a_{n+1}, a_{n+2}...)| < \epsilon## , for all ##n>N## for some N

Dick
Homework Helper
let ##a = (a_1,a_2,a_3,a_k..)## be a zero-convergent sequence.
then the sequence of sequences ##L_n = (a_1,a_2,...,a_n,0,0,0,...)## for n=1 to infinity in ##\ell## would converge to ##a##
because ##d(L_n , a) = sup|L_n - a|##, all the elements up to and including ##a_n## turns to zero, therfor ##d(L_n , a) = sup|(a_{n+1}, a_{n+2}...)|##
since for all ##n>N, d(a_n,0) < \epsilon## for some N then
##d(L_n , a) = sup|(a_{n+1}, a_{n+2}...)| < \epsilon## , for all ##n>N## for some N
Better. You should mention why there exists such an N. It's because ##a## converges to zero, right? Spelling out more reasons makes for a more readable proof.

Better. You should mention why there exists such an N. It's because ##a## converges to zero, right? Spelling out more reasons makes for a more readable proof.
okay ill add that in in my paper. but is this the whole proof or do i also need the other things? since now ##c_0## could just be a boundary around ##\ell## but i guess thats also counted as ##\ell## being dense. maybe i also need to add proof that ##\ell \subset c_0##

Dick
Homework Helper
okay ill add that in in my paper. but is this the whole proof or do i also need the other things? since now ##c_0## could just be a boundary around ##\ell## but i guess thats also counted as ##\ell## being dense. maybe i also need to add proof that ##\ell \subset c_0##
I think you've done all you need to do. But if you aren't convinced of that you should do more to convince yourself.

I think you've done all you need to do. But if you aren't convinced of that you should do more to convince yourself.
but i mean we did three things

everything from ##\ell## is in ##c_0##
all seqeunce in ##\ell## converger in ##c_0##
and all "points" in ##c_0## has a seqeunce in ##\ell## that converge to it

so its all those together right?. also thanks a lot for your help :)

Dick
Homework Helper
Sort of, we didn't prove all sequences in ##\ell## converge. We proved that if a sequence in ##\ell## converges, then it converges to a limit in ##c_0##. You're welcome!