Find the smallest value of N that will make the inequality hold for all n N.
-$$a _ { n } = \frac { 2 ^ { n } } { n ! }$$
A) L = 0, N = 8
B) L = ln 4, N = 8
C) L = ln 2, N = 6
D) diverges

Question

Find the smallest value of N that will make the inequality hold for all n > N.
-$$a _ { n } = \frac { 2 ^ { n } } { n ! }$$
A) L = 0, N = 8
B) L = ln 4, N = 8
C) L = ln 2, N = 6
D) diverges

Quizplus · Accepted Answer

For the sequence $a_n = \frac{2^n}{n!}$, as $n$ increases, the factorial in the denominator grows much faster than the exponential in the numerator. This means that $a_n$ approaches 0 as $n$ becomes very large, indicating that the sequence converges to 0. To find the smallest $N$ that makes the inequality hold for all $n > N$, we can test values to see when the sequence starts decreasing monotonically and approaches 0. Testing values, we find that after $n = 8$, the values of $a_n$ decrease more rapidly towards 0, making $N = 8$ the smallest value that satisfies the condition for all $n > N$.

Find the Smallest Value of N That Will Make the Inequality