Determine whether the given series is convergent or divergent: sqrt(n+2)/(2n^2+n+1). I thought that I should use the limit comparison test for this problem, and used (asubn) = sqrt(x+2)/(n^2+n+1) and (bsubn)= sqrt(n)/(2n^2) which simplifies to 1/(2n^(3/2). I dont know where to go from here...

Question

anonymous · Answer

so when i divided (asubn) by (bsubn) i got 2n^(3/2)*sqrt(n+2)/(2n^2+n+1)

anonymous · Answer

Any series of the form
$$
\frac 1 {n^p}, \quad p>1
$$
is convergent

anonymous · Answer

They are called p series

anonymous · Answer

You cn prove that they are convergent by the integral test

anonymous · Answer

Omg, Thats right. i totally forgot that. I was trying to go along and just simplify the problem and then plug in infinity and see what I came out with

anonymous · Answer

Great

anonymous · Answer

I could have but this problem is in the Comparison tests section, so I am practicing using those techniques.

psymon · Answer

You can still just make your comparison an actual p-series, though :3

anonymous · Answer

oh ok. I figured that even though (bsubn) is convergent because p is greater than 1, that I had to continue with the problem to show that both of them together are convergent.

psymon · Answer

Yeah, you would. You would then do your ratio of an/bn and take the limit to infinity. I just mean to say its best if you can somehow find your comparison to be a p-series when you do it.

anonymous · Answer

Im not sure how to simplify sqrt(x+2)*2n^(3/2)/(2n^2+n+1)

anonymous · Answer

Use the integral test

anonymous · Answer

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