1. ## Brief limit questions.

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

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

{ $s_{n}$} is increasing.

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

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.

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

{ $s_{n}$} is increasing.

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

$s_n$ is monotone increasing:

Induction on n:

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

Proving for $n+1$:

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

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

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.
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$s_{n+1}>s_n$

$s_n$ is bounded:

Induction on $n$:

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

Proving for $n+1$:

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

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

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

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

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

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

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

5. ## Re: Brief limit questions.

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

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

{ $s_{n}$} is increasing.

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

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

Define new sequence: $b_n$ :

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

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

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

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

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

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

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

6. ## Re: Brief limit questions.

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

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

{ $s_{n}$} is increasing.

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

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

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

CB