6
$\begingroup$

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.

  • 0
    A series which is monotone increasing is necessary bouded below. And by the first term.2015-11-28

1 Answers 1

2

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.

  • 0
    Wish I would h$a$ve thought of that $b$efore writing it all out, :( Oh well, I'll know better ne$x$t time I write a similar proof. Thanks for the input!2011-04-02