Supremum of the set of terms of the geometric series.

Question

Let $a \in \Bbb{R}$ such that $0<|a|<1$ and

$A=\{1+a+a^2+...+a^k+...+a^n|n \in \mathbb{N}\}$

Find the supremum of $A$ using the fundamental property of supremum:

$s=\sup A \Leftrightarrow s $ is an upper bound of $A$ and $\forall \epsilon>0,\exists x \in A$ such that $s-\epsilon<x$.

My thoughts are to take cases where $a$ is positive and negative and then take cases for the negative case where $n$ is an even number and odd.

My strategy is to find supremum $s$ of $A$ and then prove that any element of $A$ smaller than $s$ cannot be a supremum of $A$ contradicting the Archimedean principle:

$\forall \epsilon>0,\exists n \in \Bbb{N}$ such that $n\epsilon>1$

I have a difficulty to find the supremum of $A$ and prove that indeed is the least upper bound.

I'm new to real analysis.

I would be very thankful if someone could help me.

Thank you in advance.

Answer 1

HINT 1.

Use $\displaystyle\sum_{k=0}^n a^k={1-a^{n+1}\over1-a}$ and show that $\displaystyle\sup A={1\over1-a}$ if $a\ge0$. In other words, you must prove that, for any $\epsilon>0$: $$ {1\over1-a}-\epsilon\le{1-a^{n+1}\over1-a}\le{1\over1-a} \quad\hbox{for some $n$}. $$

HINT 2.

If $a<0$ prove that $\sup A=1$.

Answer 2

Let $b_n =\sum_{k=0}^n a^k $ and $b =\sum_{k=0}^{\infty} a^k $.

Then $b_n =\dfrac{1-a^{n+1}}{1-a} $ and $b =\dfrac{1}{1-a} $.

Then $b-b_n =\dfrac{a^{n+1}}{1-a} \to 0 $ as $n \to \infty $.

To show this:

If $a > 0$, then $a = \dfrac1{1+c}$ where $c = \dfrac1{a}-1 \gt 0$.

Therefore, by Bernoulli's inequality, $(1+c)^{m} \ge 1+mc \gt mc =m(\dfrac1{a}-1) $ so $\dfrac{a^{n+1}}{1-a} =\dfrac{1}{(1-a)(1+c)^{n+1}} \lt \dfrac{1}{(1-a)(n+1)(\dfrac1{a}-1)} = \dfrac{a}{(1-a)^2(n+1)} \to 0 $.

If $a < 0$ then $|b-b_n| =|\dfrac{a^{n+1}}{1-a}| =\dfrac{|a^{n+1}|}{1+|a|} =\dfrac{|a|^{n+1}}{1+|a|} \to 0 $ the same as for $a > 0$.