MHB Which tests should I use for convergence?

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SUMMARY

The discussion focuses on determining convergence for two series: $\displaystyle \sum_{n=2}^{\infty} (-1)^{n} \frac{n^{3} + 3n}{n^{2} + 7n}$ and $\displaystyle \sum_{n=2}^{\infty} \frac{\ln{n^3}}{n^2}$. The first series requires evaluating the limit $\displaystyle \lim_{n \rightarrow \infty} (-1)^{n} \frac{n^{3} + 3n}{n^{2} + 7n}$ to check for convergence. The second series is suggested to be analyzed using the Comparison Test, which is a standard method for assessing convergence of series.

PREREQUISITES
  • Understanding of series convergence tests, specifically the Comparison Test.
  • Familiarity with limits and their evaluation in the context of series.
  • Knowledge of alternating series and their properties.
  • Basic logarithmic functions and their behavior in series.
NEXT STEPS
  • Study the Alternating Series Test for convergence of alternating series.
  • Learn about the Comparison Test in detail, including examples and applications.
  • Explore the concept of limits in the context of series, focusing on techniques for evaluating limits.
  • Investigate the behavior of logarithmic functions in series, particularly in relation to convergence.
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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?
 
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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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