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Let the sequence $s_n$ be defined by $s_1$ = 1, $s_{n+1}$ = $\frac{1}{4} (2s_n + 5)$ for n $\in $ $\mathbb{N}$. The following is my proof and below it, my concerns.
Proof
(1) We will prove $s_n$ is increasing by induction.
That is $s_n \leq s_{n+1}$ for all n $\in \mathbb{N}$.
Since $$s_1 = 1 < s_2 = \frac{1}{4}(2(1)+5) = \frac{7}{4}$$
Now assume $s_k$ $\leq $ $s_{k+1}$
Then $$s_{k+2} = \frac{1}{4} (2s_{k+1} + 5) \geq \frac{1}{4} (2s_k +5) = s_{k+1}$$
Therefore, $s_n$ is a monotone sequence because it is increasing.

(2a) We will prove $s_n$ is bounded above.
That is, find $M \in \mathbb{R}$ such that $s_n \leq M$ for every $n \in \mathbb{N}$.
Now, we try to prove $20$ is an upper bound.
Since $s_1 = 1 < 20$
Suppose $s_k < 20$, then $$s_{k+1} = \frac{1}{4}(2s_k + 5) < \frac{1}{4}(2(20) + 5) = \frac{45}{4} < 20. $$ Hence, $s_n < 20$ for every n $\in \mathbb{N}$.
Therefore, $s_n$ is bounded above.

(2b) We will prove $s_n$ is bounded below.
That is, find $M \in \mathbb{R}$ such that $s_n \geq M$ for every $n \in \mathbb{N}$.
Now, we try to prove $0$ is a lower bound.
Since $s_1 = 1 > 0$
Suppose $s_k > 0$, then $$s_{k+1} = \frac{1}{4}(2s_k + 5) > \frac{1}{4}(2(0) + 5) = \frac{5}{4} > 0. $$
Hence, $s_n > 0$ for every n $\in \mathbb{N}$.
Therefore, $s_n$ is bounded below.
Therefore $s_n$ is bounded.

Concerns
I want to confirm I did my Induction correctly. However, my main concern is in 2b.

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    $\begingroup$ I think this looks right. For (1), I think you meant to reverse the inequality sign, and the second instance of $s_{k+2}$ should be $s_{k+1}$. Also, I think (1) and (2a) imply (2b) immediately without induction, if $s_1=1$ and $(s_n)$ is increasing, then surely it will never dip down to $0$, else it would decrease at some point. $\endgroup$ – yunone Apr 2 '11 at 2:17
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    $\begingroup$ In (1) there is a small typo, I believe, you meant s_{k+1} when you put s_{k+2}=... $\endgroup$ – WWright Apr 2 '11 at 2:21
  • $\begingroup$ Awesome! Thanks, I would have never saw those errors. $\endgroup$ – enlgmatlc Apr 2 '11 at 2:21
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    $\begingroup$ Looks much better, also, +1 for showing a good amount of work before posting. $\endgroup$ – yunone Apr 2 '11 at 2:26
  • $\begingroup$ @enlgmatlc: Punctuation goes outside in-line formulas, but inside display formulas. So $formula$., but $$formula.$$ $\endgroup$ – Arturo Magidin Apr 11 '11 at 3:01
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There are a couple of typos in your proof of (1). What you intended to write is $$s_{k+2}=\frac{1}{4}(2s_{k+1}+5) > \frac{1}{4}(2s_{k}+5)=s_{k+1}$$

There are also a couple of places where $\le$ is used, where it would be better to use $<$.

As to the sequence being bounded below, there is really nothing to do. You have proved that the sequence is increasing. It starts at 1, so 1 is a lower bound.

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  • $\begingroup$ Wish I would have thought of that before writing it all out, :( Oh well, I'll know better next time I write a similar proof. Thanks for the input! $\endgroup$ – enlgmatlc Apr 2 '11 at 2:27

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