Which of the following series is divergent? Few Pre-calc Questions

Here is the link to 4 questions on this practice quiz I have: http://oi52.tinypic.com/2mydrac.jpg

I couldn't write the questions in here because of certain symbols. If you have time, I would greatly appreciate any help you can offer. The questions shouldn't be too difficult, but I am having trouble understanding them.

a)

lim (n^3 - 5)/( n^2 ) =====> large in charge
n--> ∞

lim (n^3)/(n^2)
n--> ∞

lim n = ∞ divergent
n--> ∞

so it does not exist

Extra Example, let's find the lim of (n^3 - n^2 - n - 5)/( n^2 - n - 5) as n--> ∞

lim (n^3 - n^2 - n - 5)/( n^2 - n - 5) ==> large in charge
n--> ∞

lim (n^3)/(n^2)
n--> ∞

lim n = ∞
n--> ∞

THIS ONLY WORK IF n--> ∞ IF JUST ONLY n--> ∞
================

b)

the divergent is 1 + (3/2)^2 + (3/2)^4 + (3/2)^6

since (3/2) > 1 , it diverges by the geometric series which is increasing. all of the optional answer are convergent because of (1/2) < 1 which is decreasing to 0 .

Note: any SERIES has to approach 0 in order to converge

=================

c)

4
Σ 3^(k - 1)
k = 1

3^(0) + 3^(1) + 3^(2) + 3^(3)
1 + 3 + 9 + 27 = 40

==============

d)

since we are ending until 101, our series answer will be as :


25
Σ (4k + 1)
k = 1

let's try when k = 1 =====> (4 * 1+ 1 = 4 + 1 = 5) true
let's try when k = 25 =====> (4 * 25+ 1 = 100 + 1 = 101) true