# Convergent sequence property and proving divergence

1. Mar 18, 2012

### Sun God

I feel like I'm missing something obvious, but anyway, in the text it states:

lim as n→∞ of an+bn = ( lim as n→∞ of an ) + ( lim as n→∞ of bn )

But say an is 1/n and bn is n. Then the limit of the sum is n/n = 1, but the lim as n→∞ of bn doesn't exist and this property doesn't work...

Second thing, I was reading an example of how to prove a sequence is divergent, specifically the sequence [(-1)n], and the text proves it by contradiction. "Take any positive number $\epsilon$<1. Then the interval (a - $\epsilon$, a + $\epsilon$) has length less than 2. Therefore, it is not possible to have both odd and even terms of sequence in this interval, therefore the sequence [(-1)n] cannot converge to a."

Why does not being able to possess both odd and even terms in that interval mean that the sequence doesn't converge? It seems like a random leap of logic to me...

2. Mar 18, 2012

### jgens

The theorem should properly be written as "If $\lim_{n \to \infty}a_n$ and [/itex]\lim_{n \to \infty}b_n[/itex] exist, then $\lim_{n \to \infty} (a_n + b_n) = (\lim_{n \to \infty} a_n) + (\lim_{n \to \infty} b_n)$.

Now notice $n < n + n^{-1}$. So $n + n^{-1}$ diverges as $n \to \infty$.

A sequence $\{x_n\}_{n \in \mathbb{N}}$ converges to $x$ if and only if for every open neighborhood $U$ of $x$, there exists an $N \in \mathbb{N}$ such that $x_n \in U$ whenever $N \leq n$. The example shows that regardless of which $x \in \mathbb{R}$ you choose, there is an open neighborhood of $x$ which does not contain infinitely many of the $x_n$. This means that $\{x_n\}_{n \in \mathbb{N}}$ does not converge.

3. Mar 18, 2012

### Sun God

Thanks for the prompt response!

Okay, so to try to simplify it/put it in more understandable phrasing (for my own sake): for a sequence to be proven divergent, I just have to show that for any x there is an open neighborhood of x such that the sequence does not have infinitely many terms contained in that neighborhood?

4. Mar 18, 2012

### jgens

That is correct. More explicitly a sequence $\{x_n\}_{n \in \mathbb{N}}$ converges to $x$ if and only if for all $0 < \varepsilon$ there exists $N \in \mathbb{N}$ such that $N \leq n$ implies $|x_n - x| < \varepsilon$. This means that the sequence $\{x_n\}_{n \in \mathbb{N}}$ does not converge to $x$ if and only if there exists $0 < \varepsilon$ such that for all $N \in \mathbb{N}$ there is some $N \leq n$ such that $\varepsilon < |x_n - x|$. This is all just the standard $\varepsilon,\delta$ stuff that you learn in basic calculus classes.

Now let's look at the non-convergent case again. What it is saying is that we can find an $0 < \varepsilon$ such that no matter how big we choose $N$, we can always find an $N \leq n$ with $x_n \notin (x-\varepsilon,x+\varepsilon)$. From here it is pretty easy to work out this the open neighborhood $(x-\varepsilon,x+\varepsilon)$ does not contain infinitely many of the $x_n$.