MHB Question about determining if the series is convergent or divergent

[shamieh]

Determine if the positive term series is convergent or divergent
$$
\sum^{\infty}_{n = 1} \frac{n + cosn}{n^3 + 1}$$

can't I just ignore the cosn and look at it like this:

$$\sum^{\infty}_{n = 1} (-1)^n \frac{n}{n^3 + 1}$$

Then can't I just look at it as n--> $$\infty$$ and see that I end up with $$\frac{1}{n^2}$$ essentially and then say that it converges by the P SERIES

 

[Ackbach]

Determine if the positive term series is convergent or divergent
$$
\sum^{\infty}_{n = 1} \frac{n + cosn}{n^3 + 1}$$

can't I just ignore the cosn and look at it like this:

$$\sum^{\infty}_{n = 1} (-1)^n \frac{n}{n^3 + 1}$$

Not exactly. The $\cos(n)$ is added to the $n$ in the numerator, not multiplied. You could say this:
$$\sum_{n=1}^{\infty} \frac{n+\cos(n)}{n^3+1}\le \sum_{n=1}^{\infty} \frac{n+1}{n^3}= \sum_{n=1}^{\infty}\frac{1}{n^2}+\sum_{n=1}^{\infty}\frac{1}{n^3}.$$

 

[shamieh]

I see. So I can say:

$$0 \le \frac{n + cosn}{n^3 + 1} \le \frac{n+1}{n^3} $$

So I know that as n--> infinity for my $$\sum b_n$$ I would obtain $$\frac{1}{n^2}$$ and I know this converges by p series because $$p > 1$$ .

So by using comparison test since $$b_n$$ converges then I know $$a_n$$ converges

 

[Ackbach]

I see. So I can say:

$$0 \le \frac{n + cosn}{n^3 + 1} \le \frac{n+1}{n^3} $$

So I know that as n--> infinity for my $$\sum b_n$$ I would obtain $$\frac{1}{n^2}$$ and I know this converges by p series because $$p > 1$$ .

So by using comparison test since $$b_n$$ converges then I know $$a_n$$ converges

Right. Looks good to me!

 

[shamieh]

Thank you for your guidance Ack