proove series question..

transgalactic · #1 $Old$ December 2nd, 2008, 03:34 PM

the question in this link:
http://img372.imageshack.us/img372/7929/76106583tn6.gif

this question is obvious
its common sense

if a(n) converges to a

then if we take the function who takes the biggest member
of course it will pick the closest to "a"

it take the largest member from a1 to an

for example:
a1=6 a2=5 a3=6 a4=1 a5=8 a6=7

b1=6 b2=6 b3=6 b4=6 b5=8 b6=8

i dont know how to transform these word into math

??

the definition of convergens states that
if a sequence is bounded and monotonic then it converges

An->a
Bn=max{till An }

transgalactic · #2 $Old$ December 2nd, 2008, 11:29 PM

anyone?

transgalactic · #3 $Old$ December 3rd, 2008, 02:59 PM

i could say that if An s monotonic and converges then An+1 also converges to "a"

so Bn ->a
but here i cant say that An is monotonic

??

Mathstud28 · #4 $Old$ December 3rd, 2008, 05:51 PM

Since no one else is trying this I will give a suggestion. This is not a full solution but merely a possible building block for you to work off of. Because of the neccessary montonicity of $a_n$ why not consider three cases: $a_n<a_{n+1}\cdots$ , $a_n>a_{n+1},\cdots$ , or $a_n=a_{n+1}=\cdots$

Case Three needs no explanation.

If case two is true then obviously $\max\left\{a_n,a_{n+1},\cdots\right\}=a_n\to{a}$

And if case one is the case then we would have that $\max\left\{a_n,a_{n+1},\cdots{a_{n^2}}\right\}=a_{n^2}$

So now we just need to use the fact that $a_n\to{a}$ to show that $a_{n^2}\to{a}$

By the definition of $a_n\to{a}$ we have that for every $\varepsilon>0$ there exists a $N$ such that $N\leqslant{n}$ implies $d\left(a_n,a\right)<\varepsilon$

So now because of the above we have that $d\left(a_{n^2},a\right)<\varepsilon$ whenever $N\leqslant{n^2}\implies{\lceil{\sqrt{N}\rceil\leqslant{n}}}$ so $a_{n^2}\to{a}$

transgalactic · #5 $Old$ December 4th, 2008, 02:04 AM

its seems like a fool proof what else do i need to prove here?

case1: An<An+1
case2: An>An+1
case3: An=An+1

if case1 is true:
then MAX will pick the member with the biggest n which is An^2
and you showed how to prove that if An->0 then An^2->0

if case 2 is true:
it will pick An and by definition An goes to 0

if case3:
then no matter what it will pick because its the same.
and all the members are equaled to An
so they all go to 0.
so it lookes pretty much complete
what do i miss??

Mathstud28 · #6 $Old$ December 4th, 2008, 02:34 PM

Quote:

Originally Posted by transgalactic $View Post$

its seems like a fool proof what else do i need to prove here?

case1: An<An+1
case2: An>An+1
case3: An=An+1

if case1 is true:
then MAX will pick the member with the biggest n which is An^2
and you showed how to prove that if An->0 then An^2->0

if case 2 is true:
it will pick An and by definition An goes to 0

if case3:
then no matter what it will pick because its the same.
and all the members are equaled to An
so they all go to 0.
so it lookes pretty much complete
what do i miss??

If my proof is sufficient enough for you then I suppose you just need to rephrase it in your own words.