Determine whether the following series converge or diverge

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Homework Help Overview

The discussion revolves around determining the convergence or divergence of a given series, specifically the series defined by the sum from k=1 to infinity of (k+4)/(k^2 - 3k + 1). Participants are exploring various approaches to analyze the series.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the impact of ignoring initial terms of the series and question whether this affects convergence. There are attempts to analyze the series by dividing terms and considering limits as k approaches infinity. Some participants suggest using the comparison test to evaluate convergence.

Discussion Status

The discussion is ongoing, with participants sharing different perspectives on the convergence of the series. Some guidance has been offered regarding the comparison test and the behavior of terms as k increases, but no consensus has been reached on the conclusion.

Contextual Notes

There is a mention of the potential confusion regarding the manipulation of terms and the assumptions made about the series' behavior at infinity. Participants are navigating through these assumptions without resolving them fully.

teng125
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Determine whether the following series converge or diverge.

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

pls help...
thanx...
 
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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?
 
i try to divide everything by k and let k to infinity and finally got infinity which is diverges
 
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:
[itex]k^{2}-3k+1\leq{k^{2}-3k+3k}=k^{2}[/itex]
Divergence of your series is easily established by this result.
 
teng125 said:
i try to divide everything by k and let k to infinity and finally got infinity which is diverges
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 [itex]\frac{1+ \frac{4}{k}}{k- 3+\frac{1}{k}}[/itex].
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 [itex]\frac{k}{k^2}= \frac{1}{k}[/itex]. Does the series [itex]\Sigma\frac{1}{k}[/itex] converge or diverge? Can you use that to determine whether your original series converges or diverges?
 
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