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There is a sentence in math that says that if lim(as n->infinity)of An=L than the lim(as n->infinity) of the arithmetical average of the numbers in An is also L.

The opposite is not true.
If i am asked to show the opposite is not true all i need to do is give an example when it isn't true?or do I need to give some sort of general proof?

If I need a general proof can anyone roughly explain the way to prove it?

You are given a proof "if A, then B" (A ==> B). You want to show that the opposite "if B, then A" (B ==> A) is not true. You can do that by finding a B such that A is false; in other words, a counter-example. If (and only if) you cannot think of such an example you must prove "if B then not A" in general.

Well i can find an example but how would i go about proving this in general?any ideas?

An example would be (-1)^n when n goes from 1 to infinity.

You know about induction?

Master-Athmos wrote:You know about induction?

Yes.
How can it help me in this case?dont think it can.

First of all I think you should be explicit about what you're trying to disprove, ie what the inversion of the theorem (not sentence) really is.

Is capital A a function? I don't understand the notation An if so. I am used to f(n).

Try playing with the value of S.

CarRepairer wrote:Is capital A a function? I don't understand the notation An if so. I am used to f(n).

Guess what he means is A_n (in tex-notation)

Auswaschbar wrote:

CarRepairer wrote:Is capital A a function? I don't understand the notation An if so. I am used to f(n).

Guess what he means is A_n (in tex-notation)

Kids these days and their tex. In my day we used mechanical pencils.

An is a sequence.a[n]

I guess you were waiting for an answer... I'm interpreting from your wording that you're given the fact that the if first formula converges to a limit then the second converges to the same limit. Your job is just to prove that if the second formula converges to a limit then the first formula won't necessarily converge to the same limit (it might converge to a different limit or diverge). You gave the example of a[n] = -1^n which converges to a limit in the second but diverges in the first (from a glance anyway). That counterexample is all you need to prove what was asked.

spring forum is a great place to get your homework done.

CarRepairer wrote:I guess you were waiting for an answer... I'm interpreting from your wording that you're given the fact that the if first formula converges to a limit then the second converges to the same limit. Your job is just to prove that if the second formula converges to a limit then the first formula won't necessarily converge to the same limit (it might converge to a different limit or diverge). You gave the example of a[n] = -1^n which converges to a limit in the second but diverges in the first (from a glance anyway). That counterexample is all you need to prove what was asked.

But what if i wanted to prove it in a general form.how would i go about doing that?

You can't really prove it in 'general', just because there ARE cases when it's true (so any 'general' proof would be false just because of them). Your example of -1^n is proof enough, you don't need more.

Idea is this: if you want to prove some assumption is false, you only need 1 example of it being false. If you want to prove an assumption is true however, THEN you need a general proof.

(at least those are based on my memories of math courses, that was some years ago so I may be missing something)

Well the definitions of limits use stuff like epsilon so maybe i need to choose a certain epsilon and show there is a contradiction in some way...im just not sure how.This really calls for someone that remembers differential and integral math.

state problem exactly - copypasta problem definition with right notation and typset. jpg screenie of the homework sheet the teacher emailed out is fine.

Its all in hebrew...
Ill try to be precise.
I need to show that if the limit of the arithmetical average((a1,a2,a3....an)/n) of the numbers in a general sequence({an} is L(when n runs to infinity) it does not mean that the limit of an(when n runs to infinity) is also L.

Counterexample - an=sin(n). L = limit of (Σai)/n as n goes to infinity = 0.
The limit of an as n goes to infinity does not exist, so it's not equal to L.