91 1 point Use the ratio test to determine whether 270 conve

9\'1 (1 point) Use the ratio test to determine whether 270 converges or diverges (a) Find the ratio of successive terms. Write your answer as a fully simplified fraction. For n 7, 9001 an 0 (b) Evaluate the limit in the previous part. Enter oo as infinity and -oo as-infinity. If the limit does not exist, enter DNE. nan (c) By the ratio test, does the series converge, diverge, or is the test inconclusive?

Solution

We have given summation of (n=7 to infinity)(9ⁿ/(9n)²)

a) Let a_n=9ⁿ/(9n)² and an+1=9ⁿ⁺¹/(9(n+1))²

using the Ratio test

lim _n-->infinity |a_n+1/a_n|=lim _n-->infinity |(9ⁿ⁺¹/(9(n+1))²)/(9ⁿ/(9n)²)|

=lim _n-->infinity |(9ⁿ⁺¹/(9(n+1))²)*(9n)²/9ⁿ|

=lim _n-->infinity |(9ⁿ⁺¹(9n)²)/(9ⁿ(9(n+1))²)|

=lim _n-->infinity |9n²/(n+1)²| since common 9^n and 9^2 get canceled

lim _n-->infinity |a_n+1/a_n| =lim _n-->infinity |9n²/(n+1)²|

b) we have lim _n-->infinity |a_n+1/a_n| =lim _n-->infinity |9n²/(n+1)²|

=lim _n-->infinity |9n²/(n+1)²|

9n²/(n+1)² is positive when n-->infinity

therefore |9n²/(n+1)²| =9n²/(n+1)²

=lim _n-->infinity 9n²/(n+1)²

=9*lim _n-->infinity (n²/(n+1)²)=9*lim _n-->infinity (n²/(n^2(1+1/n)²))

=9*(lim _n-->infinity (1/(1+1/n)²) since n^2 get canceled

=9*(1/1)

=9

lim _n-->infinity |a_n+1/a_n| =9>0

c) we have lim _n-->infinity |a_n+1/a_n| =9>1

By the ratio test the given series diverges