Question

(a) Determine whether the series ∑n=0[infinity]9n2+11 is convergent or divergent.

(b) Determine whether the series ∑n=0[infinity]n+20231 is convergent or divergent.
(c) Determine whether the series ∑n /n3−11 is convergent or divergent.

Answer 1

For the series ∑n=0[infinity]9n^2+11, the series is divergent. For the series ∑n=0[infinity]n+20231, the series is divergent. For the series ∑n /(n^3−11), the series is convergent.

Step-by-step explanation:

For the series ∑n=0[infinity]9n^2+11:

The series is divergent. To determine this, we can use the comparison test. Comparing the series with the harmonic series, which is known to be divergent, we have:

lim (n→∞) ((9n^2+11)/(n)) = 9

Since the limit is not 0, the series diverges.

For the series ∑n=0[infinity]n+20231:

The series is divergent. We can use the divergence test to determine this. Taking the limit of the nth term, we have:

lim (n→∞) ((n+20231)) = ∞

Since the limit is not finite, the series diverges.

For the series ∑n /(n^3−11):

The series is convergent. We can use the limit comparison test with the series ∑n⁄n^3. Taking the limit of the ratio, we have:

lim (n→∞) ((n⁄n^3−11)/(n⁄n^3)) = lim (n→∞) (1/(1−11/n^3)) = 1

Since the limit is finite and not 0, the series converges.