L.C.T. Misgraded Calulus Problem?! Need 2nd OPINION!

[karen03grae]

HEy!

I just took a Cal. II test and used the Limit comparison test on one of my series. Here it is:

infinity
_
\
/ n^2/(n^3 +sqrt(n^(9)+1))
_
n=1

sorry if it looks hard to read. I compared it to 1/n. And it diverged. The teachers said I must compare it to n^2/n^(9/2). NOw doing it this way makes the series in question converge!
THE QUESTION IS WHY CAN'T I USE 1/N TO CHECK FOR CONVERGENCE USING THE LCT? Thanx, Karen

[selfAdjoint]

\sum_{n=1}^{\infty}1/n diverges. Why are you using it to check convergence?

[karen03grae]

I know 1/n diverges. I was trying to prove that the series in question diverges also.
I was sure that the Limit Comparison Test means that the ratio of a given series over a series I choose( usually a p-series), as n-> infinity, means that both series either converge or diverge simulaneously. Provided that the ratio is greater than zero.

I got the impression that as n goes to infinity, the ratio will tell whether or not the terms of the series are behaving similarly at infinity. If we get a ratio of "1", that is ideal. That means the terms act the same at infinity. If we get "1/2" (as I did on my test), then that is still greater than zero;

So the terms of my series and the series 1/n behave in the same way at infinity! The answer should be divergent! I am going crazy because my teacher says that I must factor out the highest power of "n" to use as a test series. Nowhere in my book does it say that!

[selfAdjoint]

Notice that \lim_{n \rightarrow \infty} \frac{1}{n}= 0 . So by your test it should converge. Problem?

[karen03grae]

No. Even though the nth term of the sequence is zero, this harmonic series does not converge! One must consider the rate at which the terms are approaching infinity. They approach too slowly. That series is definitely divergent.