Question

Let \(a_{n}\) be monotone and bounded and define \(b_{n}\) by
\(b_{n} = \frac{a_{1}+a_{2}+ ... +a_{n}}{n}\). Prove that \(b_{n}\) is monotone and bounded and therefore has a limit.

Answer 1

google 'cesaro mean" you will get lots of proofs

Answer 2

What do you think it is first?

Answer 3

Don't know if you have anything to add, freckles, but be back in a few minutes :)

Answer 4

\[n b_n=\sum_{i=1}^{n}a_i\]
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you are given a1<a2<...<a_n assuming it is monotonically increasing
and you want to show b_(n+1)>b_n for all integer n>0
You should be able to do this with induction.
This part will show b_n is also monotonically increasing.
Now to show that b_n is bounded...
we also want to show if L=<a_n<=M for some numbers L and M then L1<=b_n<=M1 for some numbers L1 and M1 .
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if we can show b_n is both monotone and bounded then b_n->to some number .
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i think the induction part looks okay
i haven't looked at the bounded part yet

Answer 5

Might have to spell it out a slight bit. Ill try messing with the induction in a sec to see if anything just hits me. But does that mean we need cases for increasing and decreasing? The question simply says monotone, not monotone increasing or monotone decreasing

Answer 6

if it is monotonic then it is either increasing or decreasing
you could probably do cases
but I think it is enough to do one case
and you can you say for the 2nd case it is proved in a similar way
seems like a lot of redundant work to also do the other case is what i'm saying

Answer 7

I'll have to do it, trust me, lol. But yeah, I'll check out the induction part of it. Now I didn't add this in, but this question is coming out of a section in which Cauchy sequences were introduced. Is there perhaps something down that route that could help with this?

Answer 8

I don't think we will need the term cauchy or the definition of cauchy.
could be wrong

Answer 9

Alrighty. Wasn't sure, but I was able to answer one or two questions without it, so maybe there's just a way to do it with cauchy but not needed. Im writing up the induction stuff, so we'll see how long before I hit a wall on that.

Answer 10

So how would that look exactly? If I wanted to look at it like b_{n+1} - b_{n} > 0
Would that be:
\[\frac{ a_{1} + a_{2} + ... a_{n+1} }{ n+1 } - \frac{ a_{1} + a_{2} + ... + a_{n} }{ n }\] Just to make sure

Answer 11

Well I wrote the statement like this \[n b_n=\sum_{i=1}^{n}a_n \\ b_1=a_1 \\ 2 b_2=a_1+a_2 \\ \text{ so we know know } a_1<a_2 \text{ since we assume we assumed } a_n \text{ was increasing } \\ \text{ so } a_1+a_1<a_1+a_2 \text{ add } a_1 \text{ \to both sides } \\ 2a_1 <2 b_2 \\ \text{ but } a_1=b_1 \\ \text{ so } 2b_1 < 2b_2 \\ b_1 <b_2 \]
this is actually going to be pretty similar to the next step of the proof

Answer 12

in fact you don't need to make any assuming
just straight up proof n b_n < (n+1)b_(n+1)

Answer 13

And that would also be by induction I assume?

Answer 14

assumption*
prove*

Answer 15

http://math.kennesaw.edu/~plaval/math4381/seqcauchy.pdf
well here it is called the direct approach

Answer 16

so i guess i lied about it being called induction :p

Answer 17

or that we were doing this by induction

Answer 18

The "direct approach" seems like it may need to be done by induction, though, lol. But yeah, I supposed that would be all you need to do is what you did. Bounded is a different story, though. So I think, lol.

Answer 19

well no we aren't done with the induction part (or i mean the direct proof)

Answer 20

and yeah the direct proof kinda reminds me of induction

Answer 21

I know, i havent written that up, but I guess I was jumping ahead to the bounded to see what the idea is. I guess I could just wait until i do the induction for monotone.

Answer 22

so our function we assumed was increasing and I'm also going to assume are sequence a_n is bounded by L and M so here is what the picture looks like
still thinking about how we can use L<=a_n<=M for all n
to show for some numbers U and V we also have U<=b_n<=V