Let $(a_k)$ be a sequence of real numbers and let $b_k=\frac{a_1+a_2+\dots a_k}{k}$ for each $k\in \mathbb{N}$. Prove that if $(a_k)$ converges to $\alpha\in \mathbb{R}$, then the sequence $(b_k)$ also converges to $\alpha$.
My answer: Let $\epsilon > 0$. Since ($a_k$) converges to $\alpha$, then there exists a $N_{\epsilon1}\in\mathbb{N}$ such that $|a_k - \alpha | < \frac{\epsilon}{2}$ for all $k \geq N_{\epsilon1}$. Notice that $\frac{a_1+...+a_n - n\alpha}{k}$ converges to $0$ as $k$ approaches infinity. Thus, there exists $N_{\epsilon2}\in\mathbb{N}$ such that $|\frac{a_1+...+a_n - n\alpha}{k}|<\frac{\epsilon}{2}$ for all $k>N_{\epsilon2}$. Now, let $N= $max$ (N_{\epsilon1},N_{\epsilon2})$. Then for any $k>N$ we get,
$$\begin{array}{rcll} |b_k-\alpha| &=& |b_k - \alpha*1|&\\ &=& |\frac{a_1+...+a_k}{k} - \frac{k\alpha}{k}|&\\ &=& |\frac{a_1+...+a_n}{k} + \frac{[a(n+1)+...+a_k]}{k} - \frac{(k-n)\alpha}{k} - \frac{n\alpha}{k}|& \text{(notice that $N_{\epsilon1}\leq N<k$)}\\ &=& |\frac{a_1+...+a_n - n\alpha}{k} + \frac{[a(n+1)+...+a_k - (k-n)\alpha]}{k}|&\\ &=& |\frac{a_1+...+a_n - n\alpha}{k} + \frac{[(a(n+1) -\alpha) + ... + (a_k - \alpha)]}{k}|&\\ &=& |\frac{a_1+...+a_n - n\alpha}{k} + \frac{(a(n+1) -\alpha)}{k} + ... + \frac{(a_k - \alpha)}{k}|&\\ &\leq& |\frac{a_1+...+a_n - n\alpha}{k}| + |\frac{(a(n+1) - \alpha)}{k}| + ... + |\frac{(a_k - \alpha)}{k}|& \text{(by the triangle inequality)}\\ &=&|\frac{a_1+...+a_n - n\alpha}{k}| + \frac{|(n+1) - \alpha|}{k} + ... + \frac{|a_k - \alpha|}{k}&\\ &<& \frac{\epsilon}{2} + [\frac{\epsilon}{2}/k + ... + \frac{\epsilon}{2}/k]& \\ &=& \frac{\epsilon}{2} + \frac{\epsilon}{2}*\frac{(k-n)}{k}&\\ &<& \frac{\epsilon}{2} + \frac{\epsilon}{2}& \text{(since $k-n<k$)}\\ &=& \epsilon& \end{array}$$ And since we found such an $N$ for arbitrary $\alpha$, then $(b_k)$ converges to $\alpha$ by definition.
I'm not sure i kept all my variables straightened out. How does this look and if you have a better idea let me know!
