Given $\lim_{n\to\infty} a_n = a$ what is the limit $\lim \limits_{n \to \infty}\frac{a_n}{3^1}+\frac{a_{n-1}}{3^2}+\ldots+\frac{a_1}{3^n}$? 
Given $\lim \limits_{n \to \infty}a_n = a$ then I need to find the limit $\lim \limits_{n \to \infty} \frac{a_n}{3} + \frac{a_{n-1}}{3^2} + \frac{a_{n-2}}{3^3} + \dotso + \frac{a_1}{3^n}$.

It seems this problem can be tackled by Stolz–Cesàro theorem. Unfortunately,  I don't know how to pick $x_n$ and $y_n$.
 A: If you have showed the convergence, then $3s_{n+1}=a_{n+1}+s_n$ where
$$
s_n=\frac{a_n}{3}+\dots \frac{a_1}{3^n}
$$
Now take the limit in both sides so you have $\lim_n s_n=a/2$
A: Fix $\epsilon > 0$. Since $\displaystyle\lim_{n \to \infty}a_n = a$, there exists an $N \in \mathbb{N}$ such that $|a_n-a| < \epsilon$ for all $n \ge N$.
Let $S_n := \dfrac{a_n}{3}+\dfrac{a_{n-1}}{3^2}+\cdots+\dfrac{a_2}{3^{n-1}}+\dfrac{a_1}{3^n}$. Then, $S_{n+1} = \dfrac{1}{3}S_n+\dfrac{1}{3}a_{n+1}$ for all $n \in \mathbb{N}$.
We can rewrite this as $S_{n+1} - \dfrac{a}{2} = \dfrac{1}{3}(S_n-\dfrac{a}{2})+\dfrac{1}{3}(a_{n+1}-a)$. Then, for $n \ge N$ we have:
\begin{align}
\left|S_{n+1} - \dfrac{a}{2}\right| & = \left|\dfrac{1}{3}(S_n-\dfrac{a}{2})+\dfrac{1}{3}(a_{n+1}-a)\right|
\\
&\le \dfrac{1}{3}\left|S_n-\dfrac{a}{2}\right|+\dfrac{1}{3}|a_{n+1}-a| 
\\
&\le \dfrac{1}{3}\left|S_n-\dfrac{a}{2}\right|+\dfrac{1}{3}\epsilon.
\end{align}
Now, use induction to show that $\left|S_n-\dfrac{a}{2}\right| \le \left(\left|S_N-\dfrac{a}{2}\right|-\dfrac{1}{2}\epsilon\right) \cdot 3^{-(n-N)}+\dfrac{1}{2}\epsilon$ for all $n \ge N$.
Then, pick $N' > N$ such that $\left|\left|S_N-\dfrac{a}{2}\right|-\dfrac{1}{2}\epsilon\right| \cdot 3^{-(n-N)} \le \dfrac{1}{2}\epsilon$ for all $n \ge N'$. 
With this choice of $N'$, we have $\left|S_n-\dfrac{a}{2}\right| \le \epsilon$ for all $n \ge N'$. 
This can be done for any $\epsilon > 0$. Thus, $\displaystyle\lim_{n \to \infty}S_n = \dfrac{a}{2}$. 
