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Let $(a_n)_{n \in \mathbb{N}}$ be a convergent sequence with limit $a \in \mathbb{R}$. Show that the arithmetic mean given by: $$s_n:= \frac{1}{n}\sum_{i=1}^n a_i \tag{A.M.} $$ also converges to $a$.

I have read: arithmetic mean of a sequence converges but unfortunately the answers there don't help me much because I don't understand their substitutions and most of it all, why their substitutions seem to work.

What I know from the problem is that since $a_n$ is convergent and $\epsilon >0$ is given, I can say that:

$$\exists N_1 \in \mathbb{N}, \forall n \geq N_1: |a_n-a|<\epsilon_1 $$ I also know that since $a_n$ is convergent, it is bound, so $(a_n) < M, \ \forall n \in \mathbb{N}$

I need to show that: $$\exists N_2 \in \mathbb{N}, \forall n \geq N_2: |s_n-a|< \epsilon_2$$ I started as follows: $$ \left|s_n-a \right| = \left|\frac{1}{n}\sum_{i=1}^na_i -a\right|= \left|\frac{1}{n}\sum_{i=1}^m(a_i-a)+\frac{1}{n}\sum_{i=m+1}^n(a_i-a) \right| \\ \leq \frac{1}{n} \sum_{i=1}^m|(a_i-a)|+\frac{1}{n}\sum_{i=m+1}^n|a_i-a|$$ I believe to understand that the left sum after the $\leq$ is finite, bound and doesn't depend on the values that $n$ takes on. However, I don't understand where all the substitutions come from and make this proof so seemingly easy to complete.

Is there a general idea I can follow to complete such proofs? Because I know that the last step is to show that the given sum is smaller than $\epsilon$. I also know that I should bring the condition $a_n < M$ into place somewhere, but I don't know where.

If I choose an $n \geq N: |a_n-a|< \epsilon'$ what does that tell me about $|a_i-a|$? I know that they are seemingly the same just with a different index.

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Here is the intuition behind the proof. As you have done, we split the sum into two pieces. Each piece is controlled differently.

To control the first piece of the sum, we note that our sequence is bounded by some constant $M$, so that this sum is at most

$$M\frac{m}{n}$$

That is, $m$ terms of value at most $M$, multiplied by the $1/n$ in front.

To control the second part of the sum, we note that our sequence $a_n$ is getting close to $a$. Thus, if we choose $m$ big enough, the $|a_n - a|$ term is less than $\epsilon /2$ for $n>m$. This means that the second sum is at most

$$\frac{\epsilon}{2}\frac{n-m}{n} \leq \frac{\epsilon}{2}$$

The last observation is that if $n$ is chosen very large, we can also have

$$M\frac{m}{n} \leq \frac{\epsilon}{2}$$

Putting these together gives us our bound.

To recap: We want to split our sum into a piece with a bounded number of terms and another piece with a growing number of terms. On the piece with a bounded number of terms, we use the bound on the sequence. On the piece with a growing number of terms, we just make sure to start far out enough that the terms are small. Finally, we shrink $\frac{1}{n}$ to make both terms as small as we please.

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  • $\begingroup$ The "piece with a bounded number of terms" is just another way to call a finite (hence bounded) sum. You don't even need to use the bound on the sequence. $\endgroup$ – Philippe Malot Oct 20 '13 at 20:29
  • $\begingroup$ Thanks a lot for your effort @Isaac Solomon, I have a much better insight into this problem now. If I may ask, does the $n-m$ in the second sum come from the fact that the sum starts by $m+1$ and goes to $n$? $\endgroup$ – Spaced Oct 20 '13 at 20:56
  • $\begingroup$ Yes. There are $n-m$ terms, which is where the $n-m$ comes from. $\endgroup$ – Elchanan Solomon Oct 20 '13 at 20:58
  • $\begingroup$ much appreciated, thanks a lot. $\endgroup$ – Spaced Oct 20 '13 at 20:59
  • $\begingroup$ can we prove the opposite using same principles? that if the arthimetic average of a sequence converges then then an converges too? if yes then what are we supposed to change? $\endgroup$ – Firas Ali Abdel Ghani Nov 12 '14 at 10:04
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Hint: Use Stolz-Cesaro Lemma.


To see the direct proof, consider that if : $$ \displaystyle\left|\frac{\displaystyle\sum_{i=1}^na_i}{n}-\frac{\displaystyle\sum_{i=1}^{n-1}a_i}{n-1}\right|\\ =\displaystyle\left|\frac{\displaystyle\sum_{i=1}^{n-1}a_i}{n(n-1)}-\frac{a_n}{n}\right|\leq \displaystyle\left|\frac{\displaystyle\sum_{i=1}^{n-1}a_i}{n(n-1)}\right|+\left|\frac{a_n}{n}\right|\\ \leq \frac{\displaystyle\max_{1\le i\le n}\left|a_i\right|}{n}\\ $$ Because $a_n$ is convergent to $a$, if $a<\infty$ then RHS goes to zero as $n$ goes to infinity and therefore the LHS goes to zero too and the sequence is convergent.

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  • $\begingroup$ can we prove the opposite using same principles? that if the arthimetic average of a sequence converges then then an converges too? if yes then what are we supposed to change? $\endgroup$ – Firas Ali Abdel Ghani Nov 12 '14 at 10:04
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Take $b_n = n$ and $c_n = \sum_{i = 1}^{n} a_i$ $$ \lim_{n\to\infty} \frac{c_{n+1} - c_n}{b_{n+1} - b_n} = \lim_{n\to\infty} \frac{a_{n+1}}{1} = a $$

$s_n = \frac{c_n}{b_n}$ and $b_n \to \infty $ and above limit exists, so Stolz–Cesàro theorem ends the proof.

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