MHB Determine Convergence/Divergence: Series Answers

[ineedhelpnow]

my final is tomorrow and my instructor gave a list of questions that will be similar to the ones on the final exam and i want to see how they should be done properly. I've been working on other problems but i can't get past these ones. thanks
determine convergence/divergence

$\sum_{n=1}^{\infty} \frac{5n^2+6}{n^4+7n+6}$

$\sum_{n=1}^{\infty} cos(\frac{4n^2+5}{2n^4+6})$id really appreciate any help. thanks!

 

[evinda]

determine convergence/divergence

$\sum_{n=1}^{\infty} \frac{5n^2+6}{n^4+7n+6}$

We use the theorem:

Let $a_n,b_n$ sequences of positive numbers such that $\frac{a_n}{b_n} \to l$,where $l>0$.Then,$\sum_{n=1}^{\infty} a_n$ converges iff $\sum_{n=1}^{\infty} b_n$ converges.

$$\lim_{n \to +\infty} \frac{\frac{5n^2+6}{n^4+7n+6}}{\frac{1}{n^2}}=5>0$$

$$\sum_{n=1}^{\infty} \frac{1}{n^2} \text{ converges, therefore } \sum_{n=1}^{\infty} \frac{5n^2+6}{n^4+7n+6} \text{ converges.}$$

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determine convergence/divergence$\sum_{n=1}^{\infty} cos(\frac{4n^2+5}{2n^4+6})$

$$\frac{4n^2+5}{2n^4+6} \to 0, \text{ therefore: } \cos(\frac{4n^2+5}{2n^4+6}) \to 1 \nrightarrow 0$$

So,the series cannot converge.

 

[evinda]

my final is tomorrow and my instructor gave a list of questions that will be similar to the ones on the final exam and i want to see how they should be done properly. I've been working on other problems but i can't get past these ones. thanks

Good luck for tomorrow's exam! (Smile)

 

[ineedhelpnow]

thanks for all your help evinda! you give the best explanations :D