Home > Calculus > Ratio Test for Convergence of an Infinite Series

How do you use the ratio test to test the convergence of the series #∑ 3^n/(4n³+5)# from n=1 to infinity?

Answer 1

Cameron Alexander

The series is divergent, see the explanation.

#L=lim_(n->oo) |a_(n+1)/a_n|#

#L=lim_(n->oo) |(3^(n+1)/(4(n+1)^3+5))/(3^n/(4n^3+5))| = lim_(n->oo) |((3^n*3)/(4(n+1)^3+5))/(3^n/(4n^3+5))|#

#L=lim_(n->oo) (3(4n^3+5))/(4(n+1)^3+5) = lim_(n->oo) (12n^3+15)/(4(n^3+3n^2+3n+1)+5)#

#L=lim_(n->oo) (12n^3+15)/(4n^3+12n^2+12n+9) = lim_(n->oo) (12+15/n^3)/(4+12/n+12/n^2+9/n^3)#

#L=12/4=3 > 1 =># the series is divergent

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Paisley Johnson

To use the ratio test to test the convergence of the series ∑ 3^n/(4n³+5), you would take the limit as n approaches infinity of the absolute value of the ratio of the (n+1)th term to the nth term:

lim (n→∞) |(a_{n+1}/a_n)|

In this case, the series is:

a_n = 3^n / (4n³ + 5)

So, you would calculate:

lim (n→∞) |(3^(n+1) / (4(n+1)³ + 5)) / (3^n / (4n³ + 5))|

Simplify the expression and then calculate the limit. If the limit is less than 1, the series converges. If it's greater than 1 or if it doesn't exist, the series diverges.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.