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

• MHB
• Vali
This is a much simpler approach than trying to solve the recurrence relation directly. In summary, the conversation discusses a given sequence and its limit, where the sequence is defined recursively based on the value of the index. The conversation also explores an alternative way to write the sequence and the limit using different constants. The conversation concludes with a discussion on finding an easier method to solve the problem.
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 $r$

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

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

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 :)

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 $\lim\limits_{n\to \infty} b_n = \dfrac{3}{4}$.

## What is a sequence?

A sequence is a list of numbers that follow a specific pattern or rule. Each number in the sequence is called a term, and the position of a term in the sequence is called its index.

## What is a limit of a sequence?

The limit of a sequence is the value that the terms of the sequence approach as the index increases without bound. In other words, it is the number that the terms get closer and closer to, but may never actually reach.

## What is the formula for finding the limit of a sequence?

The formula for finding the limit of a sequence is: limn→∞ an = L, where n represents the index of the term, an represents the term itself, and L represents the limit of the sequence.

## How do you find the limit of a sequence?

To find the limit of a sequence, you can either use the formula limn→∞ an = L or you can graph the sequence and observe the trend of the terms as the index increases. If the terms appear to approach a specific value, that value is the limit of the sequence.

## What is the significance of the index in finding the limit of a sequence?

The index is a crucial element in finding the limit of a sequence because it represents the position of a term in the sequence. As the index increases, the terms of the sequence change, and the limit is determined by the behavior of the terms as the index approaches infinity.

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