Limits: [(n + 1)(n + 2)]/(2n^2)

[Mosaness]

1. Determine whether the sequence converges, and if so find it's limit.


{\frac{(n + 1)(n + 2)}{2n²}}


2. The attempt at a solution

lim _n→∞ \frac{(n + 1)(n + 2)}{2n²}

= lim _{n →∞} \frac{(n²+ 3n + 2)}{2n²}

= lim _n→∞ \frac{n²}{2n²} + \frac{3n}{2n²} + \frac{1}{n²}

= lim _n→∞ \frac{1}{2} + \frac{3}{2n} + \frac{1}{n²}

= lim _n→∞ \frac{1}{2} + lim _n→∞\frac{3}{2n} + lim _n→∞\frac{1}{n²}

Now I am not too sure about the next part, but here is how I proceeded:

lim _n→∞ \frac{1}{2} +\frac{3}{2} lim _n→∞ \frac{\frac{1}{n}}{\frac{n}{n}} + lim _n→∞ \frac{\frac{1}{n²}}{\frac{n²}{n²}}

= \frac{1}{2} + 0 + 0

∴ The limit converges to \frac{1}{2}

 

[tt2348]

Why 1/n/(n/n)... Just use 1/n , 1/n^2 goes to 0

 

[Mosaness]

Well, I just wanted to show all of my work, in case our professor is the kind who takes off small marks. Is the way I did it correct?

 

[math man]

Now I am not too sure about the next part, but here is how I proceeded:

lim _n→∞ \frac{1}{2} +\frac{3}{2} lim _n→∞\frac{\frac{1}{n}}{\frac{n}{n}} + lim _n→∞ \frac{\frac{1}{n^{2}}}{\frac{n^{2}}{n^{2}}}

Horizontal asymptote theorem says:
\lim_{x→∞}\frac{1}{x^{n}} = 0
Dividing the numerator and denominator by n and n^{2} is unneccesary. Simplifying brings you right back to where you started:
\lim_{n→∞}\frac{1}{n} = \lim_{n→∞}\frac{\frac{1}{n}}{\frac{n}{n}} = \lim_{n→∞}\frac{1}{n}
Depending on your teacher, the quoted step above may be omitted completely as it is well known to all who have a basic understanding of limits that:
\lim_{x→a}k\,f(x) = k\lim_{x→a}f(x)
Therefore, I would consider the following step unnecessary:
\lim_{n→∞}\frac{3}{2n} = \frac{3}{2}\lim_{x→∞}\frac{1}{n}
However, if your teacher is a stickler for those types of things, then I would keep it.

 

[Mosaness]

Thank you :) I was just slightly worried that I hadn't done the process correctly, I just wanted to confirm that it was done correctly. Thank you

 

[math man]

And yes, if he said "sequence" and not "series", then you did it right.

 

[Mosaness]

How would you do it by the way of series?

 

[math man]

You wouldn't solve it by series. Generally the word "convergence" is associated with series, although it works for sequences to. If he meant sequence, you did it right. If he meant series, then he is saying...
\sum_{n=0}^{∞} \frac{(n+1)(n+2)}{2n^{2}}
Asking if this series converges or diverges is a very different question from:
\lim_{n→∞} \frac{(n+1)(n+2)}{2n^{2}}

 

[Mosaness]

Well, the question was:
Write out the first five terms of the sequence and determine if it is converging, if so, find the limit:

{\frac{(n + 1)(n + 2)}{2n²}}^+∞_{n = 1}

 

[math man]

Then I think he might mean series.
\sum_{n=1}^{∞} \frac{(n+1)(n+2)}{2n^{2}} = \frac{(2)(3)}{2(1)^{2}}+\frac{(3)(4)}{2(2)^{2}}+\ ...
Those are the first two terms in the sequence, you need 3 more (just plug in n=1, n=2, n=3, ...). As for finding if the series converges, you have to prove it using a convergence test. By solving the limit from before, you have basically performed one such test called the nth term test. If the sequence never approaches 0, then the series will never taper off and will therefore diverge.

 

[Metamorphose]

I looked at the answer manual, and whether or not it converged was done using limits. The limits shown there were somewhat similar to this. Convergence testing is taught for three more sections, my guess is it has to be shown using limits.

 

[Mark44]

Well, the question was:
Write out the first five terms of the sequence and determine if it is converging, if so, find the limit:

{\frac{(n + 1)(n + 2)}{2n²}}^+∞_{n = 1}

Then I think he might mean series.

I disagree. Assuming that Mosaness copied the problem correctly, then the question was about whether the sequence converges.

\sum_{n=1}^{∞} \frac{(n+1)(n+2)}{2n^{2}} = \frac{(2)(3)}{2(1)^{2}}+\frac{(3)(4)}{2(2)^{2}}+\ ...
Those are the first two terms in the sequence, you need 3 more (just plug in n=1, n=2, n=3, ...). As for finding if the series converges, you have to prove it using a convergence test. By solving the limit from before, you have basically performed one such test called the nth term test. If the sequence never approaches 0, then the series will never taper off and will therefore diverge.

 

[Metamorphose]

she copied it correctly. I have a copy of the book

 

[math man]

The limit you did originally is a type of convergence test for the series called an nth term test. You showed that the limit is 1/2. Because it is not 0, the series diverges. However, it does seem odd that your teacher would ask you to do something that isn't in the unit. Perhaps you could check with him to clarify? If not, pick the one you think is right.

 

[Mosaness]

According to the solution manual, it does indeed converge and the limit is 1/2. I will email my prof tomm and clarify.

 

[math man]

Sorry, I don't mean to confuse anyone. If Mark44 thinks otherwise, I would go with him. I just associate the words "converge" and "sequence" with a series problem. Plus "Write out the first five terms of the sequence" is a common first step in many series questions.

 

[Mark44]

Sorry, I don't mean to confuse anyone. If Mark44 thinks otherwise, I would go with him. I just associate the words "converge" and "sequence" with a series problem. Plus "Write out the first five terms of the sequence" is a common first step in many series questions.

A sequence can converge or diverge, so "convergent" can be an apt description for a sequence.

I interpret "Write out the first five terms of the sequence" to mean exactly what it says. Whether this is commonly done in problems of series is completely irrelevant (we now know) in this problem.

math man, please try to keep you responses in alignment with the question that is posed. If a student is having difficulties understanding the concept of convergent or divergent sequences, it's a safe bet that he or she knows nothing about infinite series.