I'm trying to prove this interesting fact of convergence. Could you please verify if my proof is correct or contains logical mistake? Thank you so much!
Let $(a_n)$ be a sequence of real numbers such that $a_n \to a$ as $n \to \infty$. Then $$\frac{\sum_{i=1}^n a_i}{n} \to a$$
My attempt:
We need two lemmas:
Lemma 1: Let $(a_n)$ be a sequence of non-negative numbers such that $a_n \to 0$ as $n \to \infty$. Then $$\frac{\sum_{i=1}^n a_i}{n} \to 0$$
Proof of Lemma 1: We have $\forall \epsilon >0, \exists N_1 \in \mathbb N,\forall n>N_1: a_n< \epsilon/2$. Moreover, there exists $N_2 \in \mathbb N$ such that $(\sum_{i=1}^{N_1} a_i) / N_2 < \epsilon/2$. Let $N = \max \{N_1,N_2\}$. Then $$\forall n > N: \frac{\sum_{i=1}^n a_i}{n} = \frac{\sum_{i=1}^{N_1} a_i}{n} + \frac{\sum_{i=N_1+1}^n a_i}{n} \le \frac{\epsilon}{2} + \frac{(n-N_1)(\epsilon/2)}{n} < \epsilon$$
The claim then follows.
Lemma 2: Let $(a_n)$ be a sequence of real numbers such that $a_n \to 0$ as $n \to \infty$. Then $$\frac{\sum_{i=1}^n a_i}{n} \to 0$$
Proof of Lemma 2: We have $$\left |\frac{\sum_{i=1}^n a_i}{n} \right | \le \frac{\sum_{i=1}^n |a_i|}{n}$$
Notice that $a_n \to 0 \iff |a_n| \to 0$. By Lemma 1, we get $\frac{\sum_{i=1}^n |a_i|}{n} \to 0$ and thus $$\frac{\sum_{i=1}^n a_i}{n} \to 0$$
Come back to our main theorem. We have $$\frac{\sum_{i=1}^n a_i}{n} - a = \frac{\sum_{i=1}^n (a_i-a)}{n}$$
We have $a_n \to a \iff (a_n-a) \to 0$. The claim then follows from Lemma 2.