# Thread: Brief limit questions.

1. ## Brief limit questions.

The question is:
Prove that the sequence is monotone and bounded. Then find the limit.

$\displaystyle s_{1}$=1 and $\displaystyle s_{n+1}$=$\displaystyle \frac{1}{4}$($\displaystyle s_{n}$+5)for n $\displaystyle \in$ N.

{$\displaystyle s_{n}$} is increasing.

Are we guessing that $\displaystyle s_{n}$ $\displaystyle \leq 2$?

2. ## Re: Brief limit questions.

You need to use tex tags, not math. And your close tag needs /, not \.

3. ## Re: Brief limit questions.

Thanks for the pointer on the formatting. And I think the Lady Gaga equation at the bottom is HILARIOUS!

4. ## Re: Brief limit questions.

Originally Posted by CountingPenguins
The question is:
Prove that the sequence is monotone and bounded. Then find the limit.

$\displaystyle s_{1}$=1 and $\displaystyle s_{n+1}$=$\displaystyle \frac{1}{4}$($\displaystyle s_{n}$+5)for n $\displaystyle \in$ N.

{$\displaystyle s_{n}$} is increasing.

Are we guessing that $\displaystyle s_{n}$ $\displaystyle \leq 2$?
$\displaystyle s_1=1, s_{n+1}=\frac{1}{4}(s_n+5)$

$\displaystyle s_n$ is monotone increasing:

Induction on n:

Say holds for $\displaystyle n$: $\displaystyle s_{n+1}>s_n$

Proving for $\displaystyle n+1$:

$\displaystyle s_{n+2}>s_{n+1}$

$\displaystyle s_{n+2}=\frac{1}{4}(s_{n+1}+5)\overset{?}{>}s_{n+1 }=\frac{1}{4}(s_n+5)$

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$\displaystyle s_{n+1}>s_n$

$\displaystyle s_n$ is bounded:

Induction on $\displaystyle n$:

Say that $\displaystyle s_n<2$ for all $\displaystyle n$ is true.

Proving for $\displaystyle n+1$:

$\displaystyle s_{n+1}=\frac{1}{4}(s_n+5)<\frac{2}{4}+\frac{5}{4} =\frac{7}{4}<2$

So, $\displaystyle s_n$ is monotonic increasing and bounded, hence $\displaystyle \lim_{n\to\infty}s_n$ exists.

We denote $\displaystyle \lim_{n\to\infty}s_n=l$ where $\displaystyle l\in\mathbb{R}$.

$\displaystyle \lim_{n\to\infty}s_{n+1}=\lim_{n\to\infty}\frac{1} {4}(s_n+5)$

$\displaystyle \lim_{n\to\infty}s_{n+1}=\lim_{n\to\infty}s_n=l$

$\displaystyle l=\frac{1}{4}(l+5)$

$\displaystyle l=\frac{5}{3}$

5. ## Re: Brief limit questions.

Originally Posted by CountingPenguins
The question is:
Prove that the sequence is monotone and bounded. Then find the limit.

$\displaystyle s_{1}$=1 and $\displaystyle s_{n+1}$=$\displaystyle \frac{1}{4}$($\displaystyle s_{n}$+5)for n $\displaystyle \in$ N.

{$\displaystyle s_{n}$} is increasing.

Are we guessing that $\displaystyle s_{n}$ $\displaystyle \leq 2$?
Another solution:

$\displaystyle s_{n+1}=\frac{1}{4}s_n+\frac{5}{4}$, $\displaystyle s_1=1$

Define new sequence: $\displaystyle b_n$ :

$\displaystyle b_n:=s_n-\frac{5}{3}$

$\displaystyle b_n$ is geometric sequence with $\displaystyle q=\frac{1}{4}$ and $\displaystyle b_1=\frac{-2}{3}$.

Hence, $\displaystyle b_n=\frac{-2}{3}\frac{1}{4^{n-1}}$.

So, $\displaystyle s_n=b_n+\frac{5}{3}=\frac{-2}{3}\frac{1}{4^{n-1}}+\frac{5}{3}$

$\displaystyle s_n=\frac{-2}{3}\frac{1}{4^{n-1}}+\frac{5}{3}$

$\displaystyle \lim_{n\to\infty}s_n=\lim_{n\to\infty}\frac{-2}{3}\frac{1}{4^{n-1}}+\frac{5}{3}=$

$\displaystyle =0+\frac{5}{3}=\frac{5}{3}$

6. ## Re: Brief limit questions.

Originally Posted by CountingPenguins
The question is:
Prove that the sequence is monotone and bounded. Then find the limit.

$\displaystyle s_{1}$=1 and $\displaystyle s_{n+1}$=$\displaystyle \frac{1}{4}$($\displaystyle s_{n}$+5)for n $\displaystyle \in$ N.

{$\displaystyle s_{n}$} is increasing.

Are we guessing that $\displaystyle s_{n}$ $\displaystyle \leq 2$?
First observe that if this converges it converges to a root of:

$\displaystyle s=\frac{1}{4}(s+5)$

knowing this greatly simplifies your work (in this case it provides a better guess for the upper bound for one thing).

CB