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 $y_n$.


closed as off-topic by Winther, Shailesh, Namaste, k170, YoTengoUnLCD Dec 29 '16 at 1:57

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  • $\begingroup$ I want to find limit of $x_n$. $\endgroup$ – False Promise Dec 29 '16 at 1:09
  • $\begingroup$ We need more details. Otherwise $a_n=1$ and $b_n=1$ (which look legal) lead to $x_n=n$... $\endgroup$ – rtybase Dec 29 '16 at 1:10
  • $\begingroup$ havent defined what $x_n$ is, how do you expect to find it's limit? $\endgroup$ – Arjang Dec 29 '16 at 1:12
  • $\begingroup$ $b_1$ should probably be $1$, since $b_1 ^ 0 = 1$. Sorry, I forgot to point it out. $\endgroup$ – False Promise Dec 29 '16 at 1:12
  • 1
    $\begingroup$ Ah, now some of the comments make no sense, there is no $x_n$ anymore ... $\endgroup$ – rtybase Dec 29 '16 at 1:16

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, $\displaystyle\lim_{n \to \infty}S_n = \dfrac{a}{2}$.


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 $\lim_n s_n=a/2$


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