OVER 9000

How would a young Anon go about calculating an average (any kind) for an infinate series?

take the sum and divide by the amount of memebers.
inf/inf = 1.
There you go.

average? average=limit. find the limit of the series.

lim(n->inf, sum(i=0..n, x_i)/n)

probably equal to the normal limit whenever the series converges. but it also works for some other series like x_n:=(-1)^n.

should be 1..n

If it is a geometric series and the terms become successively smailler by a common ratio, r, simply uses the equation a/(1-r)
where a is the first term and |r|<1

It's actually quite a complicated question.

You want lim n->inf Sn/n where Sn is Sum from 1 to n of An, the sequence.


Now, to prove this we assume the limit is A, the limit of the sequence.

This means for any epsilon, there exists N such for all n > N, |An - A| < epsilon


Now imagine |Sn/n - A| = |(Sn - A*n)/n|

= |(A1 + A2 + A3....+An - A*n)/n|
= |([A1-A] + [A2-A] + [A3-A]......+ [An - A])/n|

Now, for any epsilon > 0 there is an N S.t for all n > N |An-A| < epsilon.

So |Sn/n - A| < K/n + (n-N)*epsilon/n

Where K = A1 - A + A2 - A.....+ AN - A, a constant.

Therefore, for any espilon bigger than 0, we can choose n s.t K/n < epsilon, and (n-N)/n < 1

Therefore |Sn/n - A| < 2epsilon.


There lim n->inf Sn/n = A


Where Sn is the sum from 1 to n of a sequence An, and lim n->inf An = A

TADA!

FUCK, I wrote out a proof of this, and it didn't post.


Basically if you have a sequence An - > A as n-> inf

Then Lim n-> inf Sn/n = A

where Sn = sum from 1 to n and An

Basically you prove this by considering

|Sn/n - A| = |(Sn - A*n)/n| = |(A1 - A + A2 - A + A3 - A....+An - A)/n|

Which, by using the fact that for every epsilon > 0 there exists an N s.t for all n > N |An - A| < epsilon

Then |Sn/n - A| < K/n + (n-N)epsilon/n

Where K  = A1 - A + A2 - A....+ AN - A and is a constant.


So there is an n s.t K/n < epsilon

therefore |Sn/n - A| < epsilon

But that implies Sn/n -> A


Sorry the proofs not rigorous, I'm not typing it out again

lol, turns out it did post, whoops

I wants lots and lots of some delectable pot!

Marijuana MUST be legalized.