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The title is pretty ambiguous but I'll try to clarify it here, given the following sequences:

$$a_n \rightarrow l$$ $$b_n = \frac{\sum_{k=1}^{n}{a_k}}{n}$$

I want to show $b_n \rightarrow l$.

My method (which I think is pretty dodgy) is as follows:

$$|b_n - l| = |\frac{\sum_{k=1}^{n}{a_k}}{n} - l| \le \frac{\sum_{k=1}^{n}{|a_k-l|}}{n}$$

From the convergence of $a_n$ I have that $$\forall \epsilon > 0 ,\exists N_A,\forall n \ge N_A, |a_n-l|<\epsilon$$

Listing the terms of our new series $\ge b_n$:

$$\frac{\sum_{k=1}^{n}{|a_k-l|}}{n} = \frac{|a_1-l|}{n} + \frac{|a_2-l|}{n}+ ...+\frac{|a_{N_A}-l|}{n}+...+\frac{|a_n-l|}{n}$$

Here is where I am struggling, I need to find a value for $n$ such that the first $N_A -1$ terms are sufficiently small, but I can't seem to do so. The terms after $N_A$ are fine as they are less than $\epsilon / n$ from above.

Am I anywhere on the right tracks? Or is my approach completely incorrect? I understand I could try to show the sequence of partial sums converges as well, would that method yield better results?

Governor
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1 Answers1

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Hint: You're on the right track. Observe that, if you denote $S_{N_A}=\sum_{k=1}^{N_A}|a_k-l|$, this sum is fixed w.r.t. $n$, so we may choose $n$ so that $$\frac{\sum_{k=1}^{N_A}|a_k-l|}n<N_A\, \varepsilon\iff n>\frac{\sum_{k=1}^{N_A}|a_k-l|}{N_A\,\varepsilon}.$$

Bernard
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