MHB Find Limit of Sequence $(a_{n})$: $a_{2n+1}$

[Vali]

I have the following sequence $(a_{n})$, $a_{1}=1$
$$a_{n+1}=\begin{cases}
a_{n}+\frac{1}{2} & \text{ if } n \ is \ even \\
\frac{a_{n}}{3} & \text{ if } n \ is \ odd
\end{cases}$$
I need to find $$\lim_{n\rightarrow \infty }a_{2n+1}$$

I tried something but I didn't get too far.I rewrite the sequence:$a_{1}=1$, $$a_{n+1}=\begin{cases}
a_{n}+r & \text{ if } n \ is \ even \\
q \cdot a_{n} & \text{ if } n \ is \ odd
\end{cases}$$
where $q,r\in (0,1)$ but I don't know how to write $a_{2n+1}$ with $q$ and $r$

 

[Euge]

Hi Vali,

Note $a_{2n + 1} = a_{2n} + \dfrac1{2}$ since $2n$ is even, and $a_{2n} = \dfrac{a_{2n-1}}{3}$ since $2n-1$ is odd. Hence, if $b_n := a_{2n+1}$, then $b_n = \dfrac{b_{n-1}}{3} + \dfrac{1}{2}$. Show by induction that

$$b_n = \frac{1}{3^n} + \frac{3}{4}\left(1 - \frac{1}{3^n}\right)$$

 

[Vali]

Thank you for your help!
My attempt:
$b_{n+1} = \frac13 b_n + \frac12$
$$b_{n+1} = \frac12 + \frac13 ( \frac12 + \frac13 ( \cdots (\frac12 + \frac13 b_1 ))\cdots)\\$$
$$b_{n+1} =\frac12 (1 + \frac13 + (\frac13)^2+\cdots +(\frac13)^{n-1} ) + (\frac13)^{n} b_1\\$$

So the limit is $$\frac{1}{2}(\frac{\frac{1}{3^{n}}-1}{1}\cdot (-\frac{3}{2}))+\frac{1}{3^{n}}\cdot b_{1}=\frac{1}{2}(-1\cdot -\frac{3}{2})=\frac{3}{4}$$

Is there an easier method to solve this?Or at least how to group the terms because it was pretty difficult to me to factorize there and to see what I get.
Thanks again :)

 

[Euge]

As I've mentioned, prove by induction that $b_n = \dfrac{1}{3^n} + \dfrac{3}{4}\left(1 - \dfrac{1}{3^n}\right)$. After you've done that, then noting $\dfrac{1}{3^n} \to 0$ as $n\to \infty$, it follows from this expression of $b_n$ that $\lim\limits_{n\to \infty} b_n = \dfrac{3}{4}$.