Determine whether the sequence converges or diverges. If it converges, find the limit.
An=((-2)^n)/n!
To determine whether the sequence converges or diverges, we need to consider the behavior of the terms as n approaches infinity. In this case, we have the sequence An = (-2)^n / n!.
To find the limit, we take the limit of the terms as n approaches infinity. Let's break down the steps:
Step 1: Consider the behavior of the numerator, (-2)^n, as n approaches infinity.
The numerator (-2)^n alternates between positive and negative values as n increases. This means it does not approach a specific value but oscillates between two values. Therefore, the numerator doesn't converge to a limit.
Step 2: Consider the behavior of the denominator, n!, as n approaches infinity.
The factorial function n! grows very rapidly as n increases. As n gets larger, the denominator becomes significantly larger, making the value of 1/n! approach zero. Therefore, the denominator converges to zero.
Step 3: Consider the ratio of numerator and denominator.
Since the numerator doesn't converge but oscillates between positive and negative values while the denominator converges to zero, the ratio of (-2)^n / n! also oscillates and doesn't converge to a specific value. Therefore, the sequence diverges.
In conclusion, the sequence An = (-2)^n / n! diverges.