# Determine whether the following series converge or diverge

1. Jan 14, 2006

### teng125

Determine whether the following series converge or diverge.

(infinity) sum (k=1) (k+4)/(k^2 - 3k +1)

pls help.....
thanx.......

2. Jan 14, 2006

### arildno

1. It will be easier to work with the terms from k=3, and onwards, because the two first are negative, whereas the rest are positive.
Also, discarding any finite number of terms from a series won't affect whether it converges or diverges (agreed?)

2. So, what's the ideas you've got so far?

3. Jan 14, 2006

### teng125

i try to divide everything by k and let k to infinity and finally got infinity which is diverges

4. Jan 14, 2006

### arildno

It certainly should diverge, but I'm not too sure you've proven that rigorously.

Look at your denominator; here's one way to proceed:
$k^{2}-3k+1\leq{k^{2}-3k+3k}=k^{2}$
Divergence of your series is easily established by this result.

5. Jan 14, 2006

### HallsofIvy

Staff Emeritus
That makes no sense. If you "divide everything by k", by which I presume you mean divide each term in both numerator and denominator by k, you get $\frac{1+ \frac{4}{k}}{k- 3+\frac{1}{k}}$.
As k goes to infinity, the numerator goes to one while the denominator increases without limit: that means the terms go to 0 (not infinity!) and so the series might converge but that does not prove that it does.

Try the "comparison" test. For very very large k, the highest power "dominates" so, for very large k, the terms are close to $\frac{k}{k^2}= \frac{1}{k}$. Does the series $\Sigma\frac{1}{k}$ converge or diverge? Can you use that to determine whether your original series converges or diverges?

Last edited: Jan 15, 2006