# [SOLVED] convergence of series

• Dec 9th 2008, 07:25 PM
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

Should I use ratio test ?
• Dec 10th 2008, 07:56 AM
Mathstud28
Quote:

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
• Dec 10th 2008, 08:12 AM
sisa
is for the first case bsubn=1/3n and for the second case bsubb =2/n^3
• Dec 10th 2008, 08:15 AM
megamet2000
Quote:

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
• Dec 10th 2008, 08:15 AM
Mathstud28
Quote:

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

(Yes)
• Dec 10th 2008, 08:25 AM
sisa
Quote:

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
• Dec 10th 2008, 08:35 AM
sisa
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???
• Dec 12th 2008, 11:47 AM
megamet2000
Quote:

Originally Posted by Mathstud28
(Yes)

I think I see how you got bn but whats an?
• Dec 15th 2008, 09:05 AM
megamet2000
so sisa did you solve this problem? i dont understand how you found the limits..they both look like they would converge to 0.