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

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

$$\sum_{n=1}^\infty \frac{\sqrt{n+2}}{2n^2+n+1}$$

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

Mentor

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

For large n, $\sqrt{n + 2} \approx \sqrt{n}$, and $2n^2 + n + 1 \approx 2n^2$, so the whole expression can be compared to 1/(2n3/2), which is close to what MathematicalPhysicist said.
 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. ???
 Mentor 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 $\approx$ 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.
 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?
 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.

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

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

 You understand of course that an equation and a method are completely different things?
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.
 Mentor Sure, glad to help.

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