Question about determining if the series is convergent or divergent

[shamieh]

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

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

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

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

 

[Ackbach]

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

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

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

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:

\(\displaystyle 0 \le \frac{n + cosn}{n^3 + 1} \le \frac{n+1}{n^3} \)

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

So by using comparison test since \(\displaystyle b_n\) converges then I know \(\displaystyle a_n\) converges

 

[Ackbach]

I see. So I can say:

\(\displaystyle 0 \le \frac{n + cosn}{n^3 + 1} \le \frac{n+1}{n^3} \)

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

So by using comparison test since \(\displaystyle b_n\) converges then I know \(\displaystyle a_n\) converges

Right. Looks good to me!

 

[shamieh]

Thank you for your guidance Ack