# Thread: [SOLVED] convergence of series

1. How to check if these series converges and how to show it.
1. (n-7)/(3n^2+5n-6) when n from 1 to infinity
2.(2^2+6n-7)/(3n^5+5n^3-6) when n goes from 1 to infinity

Should I use ratio test ?

2. Originally Posted by sisa
How to check if these series converges and how to show it.
1. (n-7)/(3n^2+5n-6) when n from 1 to infinity
2.(2^2+6n-7)/(3n^5+5n^3-6) when n goes from 1 to infinity
Try the limit comparison test, if $\displaystyle \sum a_n$ and $\displaystyle \sum bn$ are infinite series and $\displaystyle \lim \frac{a_n}{b_n}=\alpha>0$ then the series share convergence/divergence

3. is for the first case bsubn=1/3n and for the second case bsubb =2/n^3

4. Originally Posted by sisa
How to check if these series converges and how to show it.
1. (n-7)/(3n^2+5n-6) when n from 1 to infinity
2.(2^2+6n-7)/(3n^5+5n^3-6) when n goes from 1 to infinity

do you mean

1. $\displaystyle \sum\limits_{n = 1}^\infty {\frac{n-7}{3n^2+5n-6}}$

2. $\displaystyle \sum\limits_{n = 1}^\infty {\frac{2n^2+6n-7}{3n^5+5n^3-6}}$

just thought id help with the formatting

5. Originally Posted by sisa
is for the first case bsubn=1/3n and for the second case bsubb =2/n^3

6. Originally Posted by megamet2000
do you mean

1. $\displaystyle \sum\limits_{n = 1}^\infty {\frac{n-7}{3n^2+5n-6}}$

2. $\displaystyle \sum\limits_{n = 1}^\infty {\frac{2n^2+6n-7}{3n^5+5n^3-6}}$

just thought id help with the formatting
yes that is what I mean

7. Ok.
I calculated that in cases:
1.lim an/bn is 1 so since bn diverges to infinity, the series an diverges to infinity
2.lim an/bn is 1/3 so since bn converges to 1 an also converges to 1

Right???

8. Originally Posted by Mathstud28
I think I see how you got bn but whats an?

9. so sisa did you solve this problem? i dont understand how you found the limits..they both look like they would converge to 0.