Thread: [SOLVED] convergence of series

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 ?

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 and are infinite series and then the series share convergence/divergence

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

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.

2.

just thought id help with the formatting

Originally Posted by sisa

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

Originally Posted by megamet2000

do you mean

1.

2.

just thought id help with the formatting

yes that is what I mean

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???

Originally Posted by Mathstud28

I think I see how you got bn but whats an?

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