Hello everyone
I need to decide if some sequences converge or not and say what the limit is with a justification. If someone could go through the method with the first one:
a = (6n - 3n^3) / (2n^3 + 4n^2)
n
thank you.
Actually, the comparison test will not work here because the sequence you gave is negative. You could treat the n's as x's and split the fraction into two terms. Integrate both terms separately. Both will be positive decreasing functions, so the integral test passes, and you will have two convergence sequences added (one is negative) to each other.
Also, keep in mind that the limit of this function is -1.5, so it definitely converges!
Does that hold generally?
If i get a sequence whose every term is less than the corresponding term of a convergent sequence, is the sequence convergent?
consider (-1,-2,-3,-4,-5,-6,.....). This sequence satisfies above conditions for the sequence $\displaystyle \frac1{n^2}$ like the previous example, but is not convergent! I think we need more than that to conclude anything.
This is just evaluating limits, isnt it?I need to decide if some sequences converge or not and say what the limit is with a justification. If someone could go through the method with the first one:
a = (6n - 3n^3) / (2n^3 + 4n^2)
n
Doesn't the sequence converge to $\displaystyle -\frac32$ ??
[QUOTE
This is just evaluating limits, isnt it?
Doesn't the sequence converge to $\displaystyle -\frac32$ ??[/QUOTE]
This is what I (think) I know:
To find out if it is convergent the limit has to be found.
So, how do I find the limit? Is there a formula? How do you arrve at the figure -3/2?
Thanks
I assumed you know it,
No formula, but do you know to evaluate limits? You can also solve by sandwiching it between two sequences like $\displaystyle -\frac32$ and $\displaystyle \frac{-3n^2}{2n^2 + n}$
Or you could just evaluate $\displaystyle \lim_{n \to \infty} \frac{6- 3n^2}{2n^2 + n}$
I will give a explicit hint:
$\displaystyle \lim_{n \to \infty} \frac{\frac6{n^2}- 3}{2 + \frac1{n}}$