Limit of (-1)^n(2n^3)/(n^3+1) using Absolute Value Theorem

[Jimbo57]

Homework Statement

Find the limit of this sequence:

(-1)ⁿ(2n³)/(n³+1)

Homework Equations

The Attempt at a Solution

At first glance I would attempt to use the absolute value theorem which gives an answer of 2 as n approaches infinite. My question is, when the theorem fails ( limit =/ 0) then does that mean the sequence is divergent? I haven't seen any documentation that explicitly states a_n is divergent if the absolute value theorem doesn't hold.

 

[Mark44]

Homework Statement

Find the limit of this sequence:

(-1)ⁿ(2n³)/(n³+1)

Homework Equations

The Attempt at a Solution

At first glance I would attempt to use the absolute value theorem which gives an answer of 2 as n approaches infinite. My question is, when the theorem fails ( limit =/ 0) then does that mean the sequence is divergent? I haven't seen any documentation that explicitly states a_n is divergent if the absolute value theorem doesn't hold.

By taking the absolute value, you're essentially stripping off the factor of (-1)ⁿ, which isn't legitimate to do. If you write down 5 or 6 terms of the sequence, you should get an idea of what it is doing.

 

[Jufro]

From an analysis point of view assume that there is some limit L. Then, by definition, for any epsilon there must exist some N such that for all n>N. Take a small epsilon, less than the limit if there was no (-1)^n, and then show that the next term is larger than said epsilon.

 

[HallsofIvy]

From an analysis point of view assume that there is some limit L.

Well, that's problemlematic with this limit!

Then, by definition, for any epsilon there must exist some N such that for all n>N. Take a small epsilon, less than the limit if there was no (-1)^n, and then show that the next term is larger than said epsilon.

What I would do is divide both numerator and denominator by $n^3$. That gives
$$(-1)^n\frac{2}{1+ \frac{1}{n^3}}$$

If it were not for that "$(-1)^n$" we could immediately say that the limit, as n goes to infinity, is 2. But what does that say about the limit with that "$(-1)^n$"?

 

[Theorem.]

The question basically boils down to $\lim_{n\rightarrow\infty}(-1)^n$. Does this limit converge? It was pointed out that with this term removed, the limit converges to the value of 2. This term switches between -1 and +1 every time we increase $n$. Intuitively what does that mean?

 

[pasmith]

My question is, when the theorem fails ( limit =/ 0) then does that mean the sequence is divergent? I haven't seen any documentation that explicitly states a_n is divergent if the absolute value theorem doesn't hold.

That's because an arbitrary $(a_n)$ might still converge. If $|a_n| \to L > 0$ then the three possibilities are that $a_n \to L$, or $a_n \to -L$, or $(a_n)$ diverges, having the property that for any $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that for all $n \in \mathbb{N}$, if $n \geq N$ then either $|a_n - L| < \epsilon$ or $|a_n - (-L)| < \epsilon$.