- #1
B3NR4Y
Gold Member
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Homework Statement
1. Let ##x_n = \frac{n^2 - n}{n} ## does ##x_n## converge or diverge?
2. Let ##x_n = \frac{(-1)^n +1}{n} ## does ##x_n## converge or diverge
Homework Equations
A sequence converges if ##\forall \epsilon > 0##, ##\exists N \in \mathbb{N} ## such that ##n\geq N ## implies ## | x_n - L | < \epsilon ##
A sequence diverges if ##\exists \epsilon_{0} > 0##, ##\forall N \in \mathbb{N}##, ##\exists n_{0} \geq N ## such that ##|x_{n_{0}} - L | > \epsilon_{0} ##
The Attempt at a Solution
For one, I noticed that ##x_n = n-1 ## which obviously diverges. But as my teacher says, I got to prove it. So I take ##\epsilon_0 = \frac{1}{2}, \, \, L \in \mathbb{R}, \, \, n \geq N, \, \, N \in \mathbb{N} ##
$$
1.) \, \, | (N - 1) - L | < \frac{1}{2}, \, \, \, \, \, \, \, \, |((N+1)-1) - L | = |N-L| <\frac{1}{2}
$$
I can apply the triangle inequality to find that ##|N-1-L| - |N-1| \le 1 ##, but I can see from 1.) that ## |N-1-L| - |N-L| < 0 ## Therefore ##x_n## diverges.
Obviously I wouldn't be posting here if that satisfied me, it doesn't. ##|N-1-L| - |N-L|=(-\infty, \, 0) ## would satisfy both of those.
2.) I noticed that ##x_n = \frac{(-1)^n +1}{n} > \frac{(-1)^n}{n} ## so all I should have to do is prove ##b_n = \frac{(-1)^n}{n} ## diverges and that should be sufficient. So if we take ##\epsilon_{0} = \frac{1}{2}, \, \, L = 1 \, \, n\geq N ##, ##|b_{n} - 1| > \epsilon ##, ##b_n = 1## when n is even and ##b_n = -1## when n is odd, therefore ##|-1 -1| = |-2| = 2 > \epsilon## which proves ##b_n## is divergent therefore proving ##x_n## is divergent..
This doesn't satisfy my either, I understanding proving convergence just fine but for some reason I find proving divergence harder.