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Divergence of a sequence 
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#1
Nov812, 06:26 PM

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 [itex] a_{n} [/itex] is a sequence of real numbers, [itex] f(n) = a_{n} [/itex] and [itex] \lim_{n→∞} f(n) [/itex] does not exist, but is not equal to ∞ or ∞, does [itex] a_{n} [/itex] necessarily diverge? If [itex] a_{n} [/itex] is a sequence of real numbers, [itex] f(n) = a_{n} [/itex] and [itex] \lim_{n→∞} f(n) = ∞ [/itex], does [itex] a_{n} [/itex] necessarily diverge? These two ideas will greatly facilitate my understanding of sequence divergence. Thanks! BiP 


#3
Nov812, 08:24 PM

P: 783

[itex] \lim_{n→∞}f(n) [/itex] does not exist, yet the limit of [itex]a_{n}[/itex] converges to 0, right?? BiP 


#4
Nov812, 08:36 PM

Mentor
P: 18,346

Divergence of a sequence
The limit [itex]\lim_{n\rightarrow +\infty} f(n)[/itex] does exist and is zero. (I assume that n is always an integer)
However, if you extend f to [itex]f(x)=\sin(2\pi x)[/itex] for [itex]x\in\mathbb{R}[/itex], then the limit [itex]\lim_{x\rightarrow +\infty} f(x)[/itex] doesn't exist. 


#5
Nov812, 08:37 PM

Sci Advisor
P: 839

Edit: ninjaed 


#6
Nov812, 10:14 PM

P: 783

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


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