Convergence of the arithmetic mean

Solution 1:

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.

Solution 2:

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.