I will be using the ratio test to prove `sum_(n=1)^oo6^n/7^n`

is absolutely convergent => the serie converge =>`sum_(n=5)^oo6^n/7^n`

also converge.

Let's rewrite the serie first. `sum_(n=5)^oo6^n/7^n=sum_(n=5)^oo(6/7)^n`

Please note that in the convergence test we look at the limit of the **absolute value** of the ratio of the n+1 term to the n term. But sinve both of our terms are positive, we will not need them.

`lim_(n->oo)abs(a_(n+1)/a_n)=`

`lim_(n->oo)[(6/7)^(n+1)]/[(6/7)^n]=lim_(n->oo)(6/7)=6/7<1`

Hence the serie absolutely converge => the serie converge.

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now