Whether the sequence converges or diverges and obtain the limit if the sequence converges.

The sequence converges to the limit 0.

Given:

The sequence is an=(−3)nn! .

Definition used:

If an is a sequence and limn→∞an exists, then the sequence an is said to be converges; otherwise it diverges.

Theorem used:

If limn→∞|an| converges to zero, then limn→∞an is converges to zero. (1)

Squeeze Theorem:

If xn≤zn≤yn for n≥N and limn→∞xn=limn→∞yn=l then limn→∞zn=l . (2)

Calculation:

Obtain the limit of the sequence to investigate whether the sequence converges or diverges.

Compute limn→∞an=limn→∞((−3)nn!) .

Compute the value of (−3)nn! .

|(−3)nn!|=|(−1)n⋅3nn!|={3nn!,if n is odd3nn!,if n is even=3nn!

Note that, the factorial of n is 1⋅2⋅3...(n−1)⋅n. .

3nn!=3⋅3⋅3...3⋅3︷n times1⋅2⋅3...(n−1)⋅n=31⋅32⋅33...3n−1⋅3n≤31⋅32⋅3n=272n

Since 0≤|(−3)nn!| and |(−3)nn!|≤272n , it can be concluded that 0≤|(−3)nn!|≤272n .

Apply Squeeze Theorem (2).

limn→∞(0)≤limn→∞(|(−3)nn!|)≤limn→∞(272n)0≤limn→∞(|(−3)nn!|)≤272⋅limn→∞(1n)0≤limn→∞(|(−3)nn!|)≤272⋅00≤limn→∞(|(−3)nn!|)≤0

Therefore, limn→∞(|(−3)nn!|)=0 .

Since |(−3)nn!| converges to zero and by using Theorem (1), (−3)nn! converges to zero.

Therefore, limn→∞((−3)nn!)=0 .

Since limn→∞an exist and by using the definition of the sequence, it can be concluded that the sequence converges to the limit 0.

Therefore, the sequence is converges to the limit 0.