Brief limit questions.

**CountingPenguins** · June 29th 2011, 09:40 PM

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 $s_{n}$ $\leq 2$ ?

**Prove It** · June 29th 2011, 09:41 PM

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

**CountingPenguins** · June 29th 2011, 09:56 PM

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

**Also sprach Zarathustra** · June 30th 2011, 03:06 AM

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 $n+1$ :

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

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

.
.
.
$s_{n+1}>s_n$

$s_n$ 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 $l\in\mathbb{R}$ .

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

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

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

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

**Also sprach Zarathustra** · June 30th 2011, 04:23 AM

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

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

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

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

**CaptainBlack** · June 30th 2011, 05:16 AM

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

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