Sequence

You prove that it doesn't converge (not a joke)

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>>271882
but how?
a proof by reduction to the absurd?

>>271887
Yes, you suppose that the function converges to a limit l for every n to infinity. If the demonstration shows it doesn't, then it diverges.

>>271898
I know how to do it with Sin(n). However, the n^3 in the idea is bothering me. Any hint?

>>271899
>idea
Sinus*

>>271902
You could use the derivative to prove if it is increasing or decreasing

>>271917
I can't use the derivative with sequences. I'm restricted to the theorem.

>>271874
Wait, whut?

Sin(anything) varies between 1 and -1, and because sin(x)+sin(x+pi)=0, the sum of sin(anything) is just the sum of sin(0) to sin(anything mod 2).

>>272238
Yeah but how do you prove it using the definitions and theorems for sequences?

Construct two subsequences, one subsequence that converges against 1 and one subsequence that converges against -1.

>>272447
The n^3 bother me for that.

>>271874
The n^3 isn't really changing anything in terms of convergence/divergence.

For any real number r between -1 and 1, any natural N, and any epsilon greater than zero, you can choose n>N so that the distance between f(n) and r is greater than epsilon.

So it doesn't converge to anything.

>>272457
I know that, the problem is that with the n^3, it hard to prove it with the definiton.

What about:

Let's suppose that Lim Sin(n^3) = L

we know that |Sin(n^3)| =< 1
so for all n, we have -1 =< L \=< 1 so with Arcsin/Sin^-1 being continuous, we have

\lim_{n \to \infty} n^3 = \lim_{n \to \infty} \sin^{-1}(\sin(n^3)) = \sin^{-1}(L) =/= +\infty

Is that good?