Question

Determine if the following series converge or diverge. Write all steps and justify your answer. Answers with wrong or missing justification will receive 0 credits. (A) [infinity]Σn=1 (-1)^n n!/10^n (Β) [infinity]Σn=1 √n^4-5/n^3-2n+11

Answer 1

To determine if the series (A) and (B) converge or diverge, we can use various convergence tests. For (A), we can use the ratio test:

(A) [infinity]Σn=1 (-1)^n n!/10^n

Let's apply the ratio test:

lim(n→∞) |((-1)^(n+1) (n+1)!)/(10^(n+1))| / |((-1)^n n!)/(10^n)|

Simplifying, we get:

lim(n→∞) |(n+1)!|/|n!|

The ratio of consecutive terms simplifies to:

lim(n→∞) (n+1)

Since this limit goes to infinity, the series (A) diverges.

Now let's analyze series (B):

(B) [infinity]Σn=1 √n^4-5/n^3-2n+11

To determine convergence or divergence, we can use the limit comparison test. Let's compare it to the series 1/n:

lim(n→∞) (√n^4-5/n^3-2n+11)/(1/n)

Simplifying, we get:

lim(n→∞) n√(n^4-5)/(n^3-2n+11)

As n approaches infinity, the numerator and denominator both grow at a similar rate. Therefore, we can apply the limit comparison test with the series 1/n:

lim(n→∞) n/n = 1

Since the limit is a nonzero finite number, the series (B) also diverges.

Therefore, both series (A) and (B) diverge.