# Thread: Convergent, divergent or definitely divergent?

1. ## Convergent, divergent or definitely divergent?

Hi,
can someone please explain to me how to find out if a set is convergent, divergent, or definitely divergent. I've seen the test to do in a textbook but i don't understand it.

I'm supposed to find out what these 5 sets are, and find the limit if its convergent.
I'd appreciate it if you showed me 1 or 2 of them as an example.
Thank you

a) an = n^2(n-3)

b) 2n^3 + 4n^2 - 5 / 8n^3 - 2n

c) n^2 / n^3 - 4

d) (-n)^-3

e) n(-1)^n

2. Originally Posted by ihatemath09
Hi,
can someone please explain to me how to find out if a set is convergent, divergent, or definitely divergent. I've seen the test to do in a textbook but i don't understand it.

I'm supposed to find out what these 5 sets are, and find the limit if its convergent.
I'd appreciate it if you showed me 1 or 2 of them as an example.
Thank you

a) an = n^2(n-3)

b) 2n^3 + 4n^2 - 5 / 8n^3 - 2n

c) n^2 / n^3 - 4

d) (-n)^-3

e) n(-1)^n
Hi! I am not 100 % sure what you mean in every example. E.g. in a) do you mean $\displaystyle n^{2(n - 3)}$ or $\displaystyle n^2 (n - 3)$? Anyway, both of these are definitely divergent for $\displaystyle n \rightarrow \infty$, since their absolute values are both bigger than $\displaystyle n^2$, which is unbound and divergent.

And in b) do you mean:
$\displaystyle 2 n^3 + 4 n^2 - \frac{5}{8 n^3} - 2 n$,
$\displaystyle \frac{2 n^3 + 4 n^2 - 5}{8 n^3 - 2 n}$,
$\displaystyle 2 n^3 + 4 n^2 - \frac{5}{8}n^3 - 2 n$
or something else? The first and the last definition is definitely divergent, because they are both bigger than $\displaystyle 2 n^3$, which is definitely divergent. The second definition converges, because it can be rewritten as:
$\displaystyle \frac{2 + 4/n - 5/n^3}{8 - 2/n^2}$
For $\displaystyle n$ going to infinity all terms vanish except $\displaystyle \frac{2}{8} = \frac{1}{4}$.

c) is also convergent regardless if you mean $\displaystyle \frac{n^2}{n^3} - 4 = \frac{1}{n} - 4$ or $\displaystyle \frac{n^2}{n^3 - 4} = \frac{1/n}{1 - 4/n^3} \rightarrow \frac{0}{1} = 0$.

d) is convergent because for positive n: $\displaystyle |(-n)^{-3}| = \frac{1}{n^3} \rightarrow 0$

e) is definitely divergent because for positive n: $\displaystyle |n(-1)^n| = n \rightarrow \infty$.