1. Owlcoffee

Do you have the series pattern given?

2. anonymous

Yes

3. anonymous

I will write it now

4. anonymous

$\sum_{1}^{\infty} \frac{ 3n - 2 }{ n ^{2} + 1 }$

5. Owlcoffee

A series is nothing more than a definable sum of terms, and it usually follows a pattern, and it is written inside the Sigma. $\sum_{n=1}^{\infty} \frac{ 3n-2 }{ n^2+1 }$ we can find any term by limiting the sum to the desired term, in this case, it's 3: $\sum_{n=1}^{n=3}\frac{ 3n-2 }{ n^2+1 }$ And, this just translates to: $\sum_{n=1}^{n=3}\frac{ 3n-2 }{ n^2+1 }=\frac{ 3(1)-2 }{ (1)^2+1 }+\frac{ 3(2)-2 }{ (2)^2+1 } + \frac{ 3(3)-2 }{ (3)^2+1 }$

6. anonymous

srry cant u just take a limit and divide the whole thing with the biggest power which in this case is n^2

7. anonymous

is it right what he jsut said?

8. anonymous

i am not sure if i am right i am pretty sure he is right

9. Owlcoffee

You take the limit when you desire to know if the series is convergent or divergent. When you want to find terms, you replace them on the actual pattern.

10. anonymous

I got another question

11. anonymous

Post it here or make open another topic?

12. Owlcoffee

Post it on another topic, so it doesn't get messy.

13. anonymous

ok

14. anonymous