Potatochip911 said:
Okay I think I'm starting to understand this.
That's good news. I will not pressure you to do more proofs here, but I strongly recommend that you take this opportunity to do some exercises on this topic from any book that covers it. If you get stuck on a problem, I suggest that you start a new thread about it.
I will show you my proof of the theorem you asked about in post #1, as well as the proof of the result that no sequence has more than one limit. We will need the following lemma.
Lemma: Let x and y be real numbers. If x<y+ε for all ε>0, then x≤y.
Proof: Suppose that x<y+ε for all ε>0. We will prove that x≤y by deriving a falsehood from the assumption that this is false. So suppose that y<x. We have (x-y)/2>0 and therefore
$$x<y+\frac{x-y}{2} =\frac{x+y}{2} <\frac{x+x}{2} =x.$$ The inequality x<x is clearly a falsehood.
Theorem: If x and y are both limits of ##(x_n)_{n=1}^\infty##, then ##x=y##.
Proof:
Let ε>0. Let ##N_1## be a positive integer such that the following implication holds for all positive integers n.
$$n\geq N_1\ \Rightarrow\ |x_n-x|<\frac{\varepsilon}{2}.$$ Let ##N_2## be a positive integer such that the following implication holds for all positive integers n.
$$n\geq N_2\ \Rightarrow\ |x_n-y|<\frac{\varepsilon}{2}.$$ Define ##n=\max\{N_1,N_2\}##. We have
$$|x-y|=|x-x_n+x_n-y|\leq|x-x_n|+|x_n-y|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon.$$ Since ε is an arbitrary positive real number, this implies that |x-y|≤0. (See the lemma). Since |x-y|≥0, this implies that |x-y|=0. This implies that x-y=0. This implies that x=y.
Comment: This result is the reason why it makes sense to talk about "the" limit of a convergent sequence instead of "a" limit.
Theorem: Let x be a real number such that |x|<1. Let ##(a_n)_{n=0}^\infty## be a sequence of real numbers such that ##|a_n|<|a_{n+1}|## for all non-negative integers n, and ##\sum_{n=0}^\infty a_n x^n## is convergent. Define ##S=\sum_{n=0}^\infty a_n x^n##. For all non-negative integers N, we have
$$\left|S-\sum_{n=0}^N a_n x^n\right|<\frac{\left|a_{N+1} x^{N+1}\right|}{1-|x|}.$$
Proof: Let ε>0. Let N be a non-negative integer. Let M be a positive integer such that the following implication holds for all positive integers m,
$$m\geq M\ \Rightarrow\ \left|\sum_{n=0}^m a_n x^n-S\right|<\varepsilon.$$ Let m be an integer greater than max{M,N}. We have
\begin{align*}
&\left|S-\sum_{n=0}^N a_n x^n\right| =\left|S-\sum_{n=0}^m a_n x_n +\sum_{n=N+1}^m a_n x^n\right|
\leq \left|S-\sum_{n=0}^m a_n x_n\right| + \left|\sum_{n=N+1}^m a_n x^n\right|\\
&<\varepsilon+\left|\sum_{n=N+1}^m a_n x^n\right| \leq \varepsilon +\sum_{n=N+1}^m |a_n x^n| = \varepsilon + \left|a_{N+1} x^{N+1}\right| \left(1+\frac{|a_{N+2}|}{|a_{N+1}|}|x|+\cdots+\frac{|a_m|}{|a_{N+1}|} |x|^{m-N-1}\right)\\
&<\varepsilon +\left|a_{N+1} x^{N+1}\right|\left(1+|x|+\cdots+|x|^{m-N-1}\right) =\varepsilon +\left|a_{N+1} x^{N+1}\right| \frac{\left(1+|x|+\cdots+|x|^{m-N-1}\right)(1-|x|)}{1-|x|}\\
&=\varepsilon+\left|a_{N+1}x^{N+1}\right| \frac{1-|x|^{m-N}}{1-|x|} <\varepsilon+\left|a_{N+1}x^{N+1}\right| \frac{1}{1-|x|}.
\end{align*}
Since ##\varepsilon## is an arbitrary positive real number and N is an arbitrary non-negative integer, this implies that
$$\left|S-\sum_{n=0}^N a_n x^n\right| \leq \frac{\left|a_{N+1}x^{N+1}\right|}{1-|x|}$$ for all non-negative integers N.