- #1
sutupidmath
- 1,630
- 4
Sequences, need help!
Well there is this problem that i am struggling to proof, i think i am close but nope nothing yet. Well, the problem goes like this:
We have two sequences [tex] \ {(a_n)} [/tex] and [tex] ({b_n})[/tex], whith the feature that [tex]{a_n}<{b_n}[/tex] for every n. Also, [tex]{a_2}={b_1}, {a_3}={b_2}... [/tex] and so forth. What i need to show is that they have the same limit, that is:
[tex]\lim_{n\rightarrow infinit} {a_n}=m=\lim_{n\rightarrow infinit} {b_n}[/tex]
I started by using the definition of the limit like this,
First i assumed that the limits are not the same so i put:
[tex]\lim_{n\rightarrow infinit} {a_n}=a[/tex] and [tex]\lim_{n\rightarrow infinit} {b_n}=b [/tex], where a is not equal to b, so it must be either smaller or bigger than it.
For every epsilon, there exists some number N_1, that for every n>N it is valid that:
I a_n-a I< epsylon, and also for every epsylon there must exist some number call it N_2, that for every n>N_2 we have:
I b_n- b I< epsylon,
then i took N=max{N_1,N_2} so for n>N, both
I b_n- b I< epsylon and I a_n-a I< epsylon are valid
then after some transformations i got:
a- epsylon<a_n< a+ epsylon and
b-epsylon< b_n < b+ epsylon
let us now suppose that b<a, than this is not possible since from a theorem i could find a number call it n_o that for every n>n_o also b_n<b<a_n , which actually contradicts the very first supposition that b_n>a_n.
now let us take b>a, and also from the supposition that b_n>a_n whe have
a-epsylon<a_n<b_n<b+epsylon
and this is where i get stuck.
Any suggestions would be really appreciated.
, so from here we have that the limit of a_n would be both b and
Well there is this problem that i am struggling to proof, i think i am close but nope nothing yet. Well, the problem goes like this:
We have two sequences [tex] \ {(a_n)} [/tex] and [tex] ({b_n})[/tex], whith the feature that [tex]{a_n}<{b_n}[/tex] for every n. Also, [tex]{a_2}={b_1}, {a_3}={b_2}... [/tex] and so forth. What i need to show is that they have the same limit, that is:
[tex]\lim_{n\rightarrow infinit} {a_n}=m=\lim_{n\rightarrow infinit} {b_n}[/tex]
I started by using the definition of the limit like this,
First i assumed that the limits are not the same so i put:
[tex]\lim_{n\rightarrow infinit} {a_n}=a[/tex] and [tex]\lim_{n\rightarrow infinit} {b_n}=b [/tex], where a is not equal to b, so it must be either smaller or bigger than it.
For every epsilon, there exists some number N_1, that for every n>N it is valid that:
I a_n-a I< epsylon, and also for every epsylon there must exist some number call it N_2, that for every n>N_2 we have:
I b_n- b I< epsylon,
then i took N=max{N_1,N_2} so for n>N, both
I b_n- b I< epsylon and I a_n-a I< epsylon are valid
then after some transformations i got:
a- epsylon<a_n< a+ epsylon and
b-epsylon< b_n < b+ epsylon
let us now suppose that b<a, than this is not possible since from a theorem i could find a number call it n_o that for every n>n_o also b_n<b<a_n , which actually contradicts the very first supposition that b_n>a_n.
now let us take b>a, and also from the supposition that b_n>a_n whe have
a-epsylon<a_n<b_n<b+epsylon
and this is where i get stuck.
Any suggestions would be really appreciated.
, so from here we have that the limit of a_n would be both b and