MHB Which tests should I use for convergence?

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To determine convergence for the series, the discussion suggests applying the limit test for the first series, where the terms must approach zero for convergence. Specifically, for the series involving (-1)^{n} (n^3 + 3n)/(n^2 + 7n), the limit as n approaches infinity should be evaluated. For the second series, ln(n^3)/n^2, the Comparison Test is recommended as a suitable method. The participants emphasize the importance of checking the behavior of the series terms as n increases. Overall, both series require careful analysis to ascertain their convergence.
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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