# Divergence of a sequence

by Bipolarity
Tags: divergence, sequence
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 P: 783 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
 Mentor P: 18,346 Yes to both questions.
P: 783
 Quote by micromass 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

 Mentor P: 18,346 Divergence of a sequence 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.
Sci Advisor
P: 839
 Quote by Bipolarity $\lim_{n→∞}f(n)$ does not exist
Why do you say that?

Edit: ninjaed
P: 783
 Quote by micromass 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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