# Yet Another Convergence Question

Homework Statement
Let a_1, a_2, ... be a sequence of real numbers that converges to a. Let b_n = 1/n(a_1 + ... + a_n). Prove that b_1, b_2, ... also converges to a.

The attempt at a solution
Let e > 0. We know that there is an N such that |a_n - a| < e for all n > N. Now |nb_n - na| <= |a_1 - a| + ... + |a_n - a|, but I don't see how we can have |a_i - a| < e for i = 1, ..., n. Any tips?

Dick
Homework Helper
b_m for m>N is (1/m)*(sum(a_i for i=1 to N)+(1/m)*sum(a_i for i=N+1 to m)). The first sum is finite and doesn't depend on m. So it's contribution to the limit is 0.

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I don't understand why the contribution to the limit is 0. Does that mean that |a_i - a| < e for all i = 1, ..., N?

Dick
Homework Helper
Not at all. The point is that the contribution of the first N terms, where you don't have a bound, to the limit of the b_i is insignificant. It's only a finite number of terms. For large i>>N, b_i is the average over a huge number of terms.

Sorry for being dense. Are you're saying that the first N terms of the series do not contribute to the sum of the series? I wasn't thinking in terms of series. What would be the epsilon-argument?

Dick
Homework Helper
Of course they contribute! But you can make their contribution unimportant by looking at large enough b_m. Look b_m=(A+sum(b_i, for i>N))/m where A is the sum of the first N terms. The sum is m-N numbers all of which are within epsilon of the limit 'a'. What's the limit as m->infinity? You will have to account for A in your limit argument, but you can make A/m as small as you want by picking m large enough.

Look b_m=(A+sum(b_i, for i>N))/m where A is the sum of the first N terms.
I think you meant to write a_i instead of b_i and "for N < i <= m" instead of "for i>N".

What's the limit as m->infinity? You will have to account for A in your limit argument, but you can make A/m as small as you want by picking m large enough.
I see know. For some reason, I was thinking that A varied with m. Don't ask me why. Thanks for your help and your patience!

Dick