For the sequence below, determine whether they converge or not, and find the limit when the sequence converges.

an=4^n/n!

an=nsin(1/n)

To determine whether a sequence converges or not, we need to examine its behavior as n approaches infinity. Let's analyze the two sequences you provided:

1. Sequence: an = 4^n/n!
To determine if this sequence converges, we can find the limit of the sequence as n approaches infinity.

Step 1: Rewrite the sequence formula using exponential notation for 4^n:
an = (4/1)(4/2)(4/3)...(4/n)

Step 2: Simplify the terms in the formula:
an = (4/n)(4/(n-1))(4/(n-2))...(4/3)(4/2)(4/1)

Step 3: Notice that all terms cancel out except for 4/1, resulting in a constant value of 4.

Step 4: Therefore, the limit of this sequence as n approaches infinity is 4, which means the sequence converges to the value 4.

2. Sequence: an = n*sin(1/n)
To determine if this sequence converges, we can also find the limit of the sequence as n approaches infinity.

Step 1: Rewrite the sequence formula using the limit notation:
lim(n -> infinity) of n*sin(1/n)

Step 2: Apply the limit of trigonometric functions as their argument approaches zero:
lim(x -> 0) of sin(x) = 1

Step 3: Substitute 1/n for x in the above limit expression:
lim(n -> infinity) of n*sin(1/n) = lim(n -> infinity) of sin(1/n)/(1/n)

Step 4: Simplify the limit expression using the limit of ratios:
lim(x -> 0) of sin(x)/x = 1

Step 5: With this in mind, we have:
lim(n -> infinity) of n*sin(1/n) = 1

Therefore, the limit of this sequence as n approaches infinity is 1, which means the sequence converges to the value 1.

In summary, the first sequence converges to 4, and the second sequence converges to 1.