MHB Do we distinguish cases for n?

[evinda]

Hello! (Wave)

Let $\alpha \neq 1$. I have shown that

$$1+\alpha+ \dots+ \alpha^k+\dots+ \alpha^n=\frac{1-\alpha^{n+1}}{1-\alpha}.$$

Now I want to show that, when $0<|\alpha|<1$, then the set $\{1+ \alpha+ \dots+ \alpha^k+\dots+ \alpha^n \mid n \in \mathbb{N} \}$ is bounded. And I want to find its least upper bound.

I have shown that when $0< \alpha<1$, then the lower bound of the set is $0$ and the least upper bound is $\frac{1}{1-\alpha}$.

So it suffices to find the upper and lower bound of the set when $-1< \alpha<0$.

In this case, we cannot find a general inequality for $\alpha^{n+1}$. Do we distinguish cases for $n$ ? (Thinking)

 

[I like Serena]

In this case, we cannot find a general inequality for $\alpha^{n+1}$.

Hey evinda! (Smile)

Can't we use $-1-|\alpha|-|\alpha|^2-...$ and $1+|\alpha|+|\alpha|^2+...$ as lower and upper bounds? (Wondering)

 

[evinda]

Hey evinda! (Smile)

Can't we use $-1-|\alpha|-|\alpha|^2-...$ and $1+|\alpha|+|\alpha|^2+...$ as lower and upper bounds? (Wondering)

Ah I see... But can we also find a least upper bound for the general case? (Thinking)

 

[Euge]

Note $1/(1 - \alpha)$ cannot be the least upper bound when $-1 < \alpha < 0$. For if $\alpha = -1/2$, $1/(1 - \alpha) = 2/3$, but $1 + \alpha = 3/2 > 2/3$.

To handle the negative case, let $\beta = -\alpha$ so that $0 < \beta < 1$. Then

$$1 + \alpha + \cdots + \alpha^n = \begin{cases}\dfrac{1 - \beta^{n+1}}{1+\beta}&\text{if $n$ is odd}\\\dfrac{1 + \beta^{n+1}}{1 + \beta}&\text{if $n$ is even}\\\end{cases}$$

Since $0 < \beta < 1$, in the both cases above, the sum on the left-hand side is bounded above by $\sigma := (1 + \beta^3)/(1 + \beta) = (1 -\alpha^3)/(1 - \alpha)$. Thus, $\sigma$ is an upper bound for your set, call it $S$. Since $1 + \alpha + \alpha^2 = (1 - \alpha^3)/(1 - \alpha) = \sigma$, then $\sigma$ is in fact the least upper bound of $S$.

 

[evinda]

Note $1/(1 - \alpha)$ cannot be the least upper bound when $-1 < \alpha < 0$. For if $\alpha = -1/2$, $1/(1 - \alpha) = 2/3$, but $1 + \alpha = 3/2 > 2/3$.

To handle the negative case, let $\beta = -\alpha$ so that $0 < \beta < 1$. Then

$$1 + \alpha + \cdots + \alpha^n = \begin{cases}\dfrac{1 - \beta^{n+1}}{1+\beta}&\text{if $n$ is odd}\\\dfrac{1 + \beta^{n+1}}{1 + \beta}&\text{if $n$ is even}\\\end{cases}$$

Since $0 < \beta < 1$, in the both cases above, the sum on the left-hand side is bounded above by $\sigma := (1 + \beta^3)/(1 + \beta) = (1 -\alpha^3)/(1 - \alpha)$. Thus, $\sigma$ is an upper bound for your set, call it $S$. Since $1 + \alpha + \alpha^2 = (1 - \alpha^3)/(1 - \alpha) = \sigma$, then $\sigma$ is in fact the least upper bound of $S$.

How did you find these upper bounds?
I found that the upper bound is $\frac{1}{1+\beta}$ when $n$ is odd, and $\frac{2}{1+\beta}$ when $n$ is even. (Thinking)

I found the upper bounds as follows.We have that $\beta>0$. So $1-\beta^{n+1} <1 \Rightarrow \frac{1-\beta^{n+1}}{1+\beta}<\frac{1}{1+\beta}$.

We also have that $\beta<1 \Rightarrow \beta^{n+1}<1$. So $\frac{1+\beta^{n+1}}{1+\beta}<\frac{2}{1+\beta}$.

 

[Euge]

First, let me make a correction above: In my example where $\alpha = -1/2$, I meant to say $1/(1 - \alpha) = 2/3$, but $1 + \alpha + \alpha^2 = 1 - 1/2 + 1/4 = 3/4 > 2/3$.

Now since $0 < \beta < 1$, then when $n$ is even, $\beta^{n+1} \le \beta^{2 + 1} = \beta^3$, so then $(1 + \beta^{n+1})/(1 + \beta) \le (1 + \beta^3)/(1 + \beta)$ for even $n$. Since $\beta > 0$, then for all $n$, $(1 - \beta^{n+1})/(1 + \beta) < (1 + \beta^3)/(1 + \beta)$. This is how I obtain $\sigma := (1 + \beta^3)/(1 + \beta) = (1 - \alpha^3)/(1 - \alpha)$ as an upper bound for $S$. Since $\sigma \in S$, then $\sigma$ is the least upper bound for $S$.

Note that your bounds are not the upper bounds, as there are infinitely many upper bounds in the even and odd cases (in fact, there are infinitely many upper bounds for any nonempty bounded subset of the reals).

 

[evinda]

Now since $0 < \beta < 1$, then when $n$ is even, $\beta^{n+1} \le \beta^{2 + 1} = \beta^3$, so then $(1 + \beta^{n+1})/(1 + \beta) \le (1 + \beta^3)/(1 + \beta)$ for even $n$.

This holds for $n \geq 3$. It does not hold for $n=1$, right?

 

[I like Serena]

This holds for $n \geq 3$. It does not hold for $n=1$, right?

n=1 isn't even is it?
And doesn't it hold for n=2? (Wondering)

 

[evinda]

n=1 isn't even is it?
And doesn't it hold for n=2? (Wondering)

Oh yes, you are right.

To show that $\frac{1+\beta^3}{1+\beta}$ is indeed the least upper bound, I supposed that there is an $\epsilon>0$ such that $\frac{1+\beta^3}{1+\beta}-\epsilon$ is a smaller upper bound.

When $n$ is even, then we will have that $\frac{1+\beta^{n+1}}{1+\beta}<\frac{1+\beta^3}{1+\beta}-\epsilon, \forall n \in \mathbb{N}$.

Then we get that $n<\frac{\ln{(\beta^3-\epsilon (1+\beta))}}{\ln{(\beta)}}-1$, which is a contradiction.

When $n$ is odd, from $\frac{1-\beta^{n+1}}{1+\beta}< \frac{1+\beta^3}{1+\beta}$ we get that $\beta^{n+1}>-\beta^3+\epsilon(1+\beta)$.

Is $-\beta^3+\epsilon(1+\beta)$ positive? Or how do we find a restriction for $n$ in order to get a contradiction?

 

[Euge]

I actually gave a proof that $\sigma$ is the least upper bound in my first post, adding more details in my second post.

 

[evinda]

Since $\sigma \in S$, then $\sigma$ is the least upper bound for $S$.

How do we know that there is no other element in $S$ that is a lower upper bound of the set? (Thinking)

 

[Euge]

How do we know that there is no other element in $S$ that is a lower upper bound of the set? (Thinking)

If you have a nonempty bounded set $X$ with an upper bound $s$ such that $s\in X$, then $s$ is the least upper bound of $X$. For since $s\in X$, $s\le \sup X$. On the other hand, since $s$ is an upper bound for $X$, $s \ge \sup X$. This shows $\sup X = s$.

 

[evinda]

If you have a nonempty bounded set $X$ with an upper bound $s$ such that $s\in X$, then $s$ is the least upper bound of $X$. For since $s\in X$, $s\le \sup X$. On the other hand, since $s$ is an upper bound for $X$, $s \ge \sup X$. This shows $\sup X = s$.

Ah I see... And how can we find the lower bound of the set when $-1<\alpha<0$ ?Also.. are my lower and upper bound of the set correct for the case when $0<\alpha<1$ ?