Which tests should I use for convergence?

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Discussion Overview

The discussion centers around the convergence of two series: one involving an alternating series with polynomial terms and another involving a logarithmic function divided by a polynomial. Participants seek to identify appropriate convergence tests for these series.

Discussion Character

  • Homework-related, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant presents two series and asks which tests should be used for convergence.
  • Another participant calculates the limit of the first series' terms as \( n \) approaches infinity, questioning the behavior of the series.
  • A participant challenges the clarity of a notation used in the first series, indicating potential confusion.
  • It is noted that a necessary condition for convergence is that the terms of the series must approach zero; a participant emphasizes this point.
  • For the second series, one participant suggests using the Comparison test as a potential method for determining convergence.

Areas of Agreement / Disagreement

Participants express differing views on the appropriate tests for convergence, with no consensus reached on which specific tests to apply to the series presented.

Contextual Notes

There are unresolved questions regarding the notation and the behavior of the series terms, which may affect the choice of convergence tests.

Who May Find This Useful

Students and individuals interested in series convergence, particularly those studying calculus or mathematical analysis.

Tebow15
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Test these for convergence.

3.
infinity
E...((-1)^n)*(n^3 + 3n)/((n^2) + 7n)
n = 2

4.
infinity
E...ln(n^3)/n^2
n = 2

note: for #3: -((n^2 + 3n))/n) is all to the power of e

Btw, E means sum.

Which tests should I use to solve these?
 
Last edited by a moderator:
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cacophony said:
3. $\displaystyle \sum_{n=2}^{\infty} (-1)^{n} \frac{n^{3} + 3\ n}{n^{2} + 7\ n}$

What is $\displaystyle \lim_{n \rightarrow \infty} (-1)^{n} \frac{n^{3} + 3\ n}{n^{2} + 7\ n}$?...

Kind regards

$\chi$ $\sigma$
 
cacophony said:
note: for #3: -((n^2 + 3n))/n) is all to the power of e

It is not clear what do you mean by that ?
 
A necessary (though not sufficient) condition for a series to converge is that the terms in the series eventually have to vanish to 0. That means that if they do NOT vanish to 0, the series is divergent.

So what happens to the terms in the first series as you go along?
 
For the number 4
$$\sum_2^\infty\frac{\ln{n^3}}{n^2}$$

I would try the good old Comparison test!

Give it a try !
 
For number 4

$$\ln(n) < \sqrt{n} $$
 

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