Convergent sequence property and proving divergence

[Sun God]

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

lim as n→∞ of a_n+b_n = ( lim as n→∞ of a_n ) + ( lim as n→∞ of b_n )

But say a_n is 1/n and b_n is n. Then the limit of the sum is n/n = 1, but the lim as n→∞ of b_n 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 possesses 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...

 

[jgens]

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

lim as n→∞ of a_n+b_n = ( lim as n→∞ of a_n ) + ( lim as n→∞ of b_n )

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

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 &lt; n + n^{-1}. So n + n^{-1} diverges as n \to \infty.

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 possesses 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...

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.

 

[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?

 

[jgens]

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?

That is correct. More explicitly a sequence \{x_n\}_{n \in \mathbb{N}} converges to x if and only if for all 0 &lt; \varepsilon there exists N \in \mathbb{N} such that N \leq n implies |x_n - x| &lt; \varepsilon. This means that the sequence \{x_n\}_{n \in \mathbb{N}} does not converge to x if and only if there exists 0 &lt; \varepsilon such that for all N \in \mathbb{N} there is some N \leq n such that \varepsilon &lt; |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 &lt; \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.