Finding "a" and "b" in an infinite series limit comparison test

MillerGenuine

1. The problem statement, all variables and given/known data

[tex]
\sum_{n=1}^\infty \frac{\sqrt{n+2}}{2n^2+n+1}
[/tex]

How do I identify my a_n and my b_n?
In this particular problem you need to use the Limit comparison test which is your "a_n" divided by your "b_n". I know how to solve the problem once these variables are identified, but for each question i attempt to do, i am not seeing a pattern in how to identify your a and b.

MathematicalPhysicist

You can show that the term of this sum is smaller than:
1/( n sqrt(n+2))
which is in return smaller than 1/(n sqrt n)

and if I recall correctly the sum of 1/n^a converges for a>1.

MillerGenuine

for this problem they get
[tex]
a_n= \frac{\sqrt(n+2)}{2n^2+n+1}
b_n= \frac{1}{n^(3/2)}
[/tex]

how are these values obtained?
I realize a_n is just the initial function..but how did they get b_n?
I know to divide them, & i know how to get the answer from this point, i just have no idea how they got their "b_n" value

Mark44

For large n, [itex]\sqrt{n + 2} \approx \sqrt{n}[/itex], and [itex]2n^2 + n + 1 \approx 2n^2[/itex], so the whole expression can be compared to 1/(2n^3/2), which is close to what MathematicalPhysicist said.

MillerGenuine

Is there some kind of equation to find b_n?
Im not sure the terminology of "large n". why can you approximate n+2 to be just n, and 2n^2 + n + 1 to be just 2n^2. ???

Mark44

If you're looking for some formula that you can use instead of thinking, no, there isn't. "Large n" simply means in the limit as n goes to infinity.

For n + 2:
If n = 10, n + 2 = 12
If n = 100, n + 2 = 102
If n = 1000, n + 2 = 1002
The larger n gets, the closer n and n + 2 are, relatively speaking. The difference is always 2, but the relative difference gets smaller and smaller.

In a polynomial in n, the dominant term when n is large is the term of highest degree. That's why I can say that n + 2 [itex]\approx[/itex] n, for large n, hence their square roots are approximately equal as well.

That's also why I can say that for large n, 2n² + n + 1 is about equal to 2n². The larger n is, the smaller the effect of the lower degree terms.

MillerGenuine

So would it be safe to say that for any problem where i must find my b_n, such as the one above, i essentially take the term with the highest degree (in the numerator and denom) and drop the rest? because n becomes so large as it approaches infinity the rest of the polynomial is negligible?

MillerGenuine

When i had said "equation" in my previous post i meant a method to always find b_n. like i asked in the above post, taking the term with highest degree and dropping the rest.

Mark44

If you're dealing with a series whose general term is a rational function, yes.

You understand of course that an equation and a method are completely different things?

MillerGenuine

Of course. I wasnt sure quite how to word my question so I had to make a stretch. I tend to miss the smallest details (such as this b_n issue) yet can understand everything else magnificently. Go figure. I appreciate the help though, it makes much more sense.

Mark44

Sure, glad to help.