# Infinite series convergence II

#### cacophony

##### New member
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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#### chisigma

##### Well-known member
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

$\chi$ $\sigma$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
note: for #3: -((n^2 + 3n))/n) is all to the power of e
It is not clear what do you mean by that ?

#### Prove It

##### Well-known member
MHB Math Helper
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?

#### Barioth

##### Member
For the number 4
$$\displaystyle \sum_2^\infty\frac{\ln{n^3}}{n^2}$$

I would try the good old Comparison test!

Give it a try !

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
For number 4

$$\displaystyle \ln(n) < \sqrt{n}$$