What Methods Can Be Used to Prove Sequence Divergence?

Bipolarity
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I'm trying to understand divergence of a sequence (not series). What methods can I use to prove divergence? I know that convergence can be proven using various methods, such as squeeze theorem and sum, difference, product and quotient rule etc.

Could I use the following to prove divergence?

If a_{n} is a sequence of real numbers, f(n) = a_{n} and \lim_{n→∞} f(n) does not exist, but is not equal to ∞ or -∞, does a_{n} necessarily diverge?

If a_{n} is a sequence of real numbers, f(n) = a_{n} and \lim_{n→∞} f(n) = ∞, does a_{n} necessarily diverge?

These two ideas will greatly facilitate my understanding of sequence divergence.
Thanks!

BiP
 
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Yes to both questions.
 
micromass said:
Yes to both questions.

Hey micro, but what about the sequence a_{n} = sin(2πn). It is the case that
\lim_{n→∞}f(n) does not exist, yet the limit of a_{n} converges to 0, right??

BiP
 
The limit \lim_{n\rightarrow +\infty} f(n) does exist and is zero. (I assume that n is always an integer)

However, if you extend f to f(x)=\sin(2\pi x) for x\in\mathbb{R}, then the limit \lim_{x\rightarrow +\infty} f(x) doesn't exist.
 
Bipolarity said:
\lim_{n→∞}f(n) does not exist

Why do you say that?

Edit: ninjaed
 
micromass said:
The limit \lim_{n\rightarrow +\infty} f(n) does exist and is zero. (I assume that n is always an integer)

However, if you extend f to f(x)=\sin(2\pi x) for x\in\mathbb{R}, then the limit \lim_{x\rightarrow +\infty} f(x) doesn't exist.

micromass, I'm sorry I think I misphrased my question. When I refer to f(n) in my original post, I refer to it as a function with domain ℝ as opposed to a_{n} which I take to be defined only for natural numbers.

Given this clarification, which of the following original statements is true and why?

BiP
 
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