Here's a good sequence question...?

Question

Suppose you have a sequence of real numbers a_1, a_2, a_3... such that the sequence converges to some number a.

Let S(a_i) denote the sum of consecutive members of the sequence from i=1 to n.
Show: lim (n goes to +inf) of 1/n*S(a_i) = a.

Answer 1

In other words, show that the Cesaro limit equals the "normal" limit.

I write S[n] = (a[1] + a[2] + .. + a[n]) / n.

Let e > 0. By definition, there exists n such that for m >= n, |a[m] - a| < e.

Let L be an integer greater than both |S[n] - n a| / e and n. Suppose k >= L.

Then
|S[k] - k a| = |a[1]+...+a[n]+...+a[k] - k a|
... < |a[1]+...+a[n] - n a| + |a[n+1]+...+a[k] - (k-n) a|
... = |S[n] - n a| + |(a[n+1] - a) + ... + (a[k] - a)|
... < k e + (k - n) e
... < 2 k e
|S[k] / k - a| < 2 e

This proves that for every 2e > 0, there exists L such that for k >= L, |S[k] / k - a| < 2e. By definition this means that a is the limit of S[k] / k.

Answer 2

Here's how I like to do this. We know that |a_n - a|0 by simply taking n>N. Consider the following (
|(1/n)(a_1+...+a_n)-(1/n)(na)|
Let N_1 be such that for all n>N_1, |a_n - a|
(1/n)[|a_1 - a|+...+|a_{N_1} - a|] and (1\n)[|a_{N_1 + 1} - a|+...+|a_n - a|]. For n>N_1, the second term is ((n-N_1)/n)(e/2)
The other has a finite numerator and a decreasing denominator, so there exists an N_2 so that for n>N_2, (1/n)[|a_1 - a|+...+|a_{N_1} - a|]N, each term is
Now think of the reverse direction: If lim (n goes to +inf) of 1/n*S(a_i) = a, does that imply {a_n} converges to a? Or even converges at all? It turns out the answer is no, since the sequence a_n = (-1)^n is clearly divergent and yet for this sequence, lim (n goes to +inf) of 1/n*S(a_i) = 0.

Answer 3

i don't think this is sequential enough...