Use the Ratio Test to determine the convergence of: [tex]$\sum_{n=1}^{\infty} \frac{3^n}{n!}$[/tex]
Answer (2)
Apply the Ratio Test by computing the limit of the ratio of consecutive terms: L = lim n → ∞ a n a n + 1 .
Determine a n + 1 for the given series a n = n ! 3 n , which is a n + 1 = ( n + 1 )! 3 n + 1 .
Simplify the ratio a n a n + 1 to n + 1 3 .
Compute the limit L = lim n → ∞ n + 1 3 = 0 , and since L < 1 , the series converges by the Ratio Test.
Explanation
Problem Analysis We are asked to determine the convergence of the series ∑ n = 1 ∞ n ! 3 n using the Ratio Test. The Ratio Test is a powerful tool for determining the convergence or divergence of an infinite series. It involves examining the limit of the ratio of consecutive terms in the series.
Ratio Test The Ratio Test states that for a series ∑ n = 1 ∞ a n , we compute the limit: L = n → ∞ lim a n a n + 1 If L < 1 , the series converges absolutely. If 1"> L > 1 , the series diverges. If L = 1 , the test is inconclusive.
Finding a_{n+1} In our case, a n = n ! 3 n . So, we need to find a n + 1 , which is obtained by replacing n with n + 1 in the expression for a n : a n + 1 = ( n + 1 )! 3 n + 1
Computing the Ratio Now, we compute the ratio a n a n + 1 : a n a n + 1 = n ! 3 n ( n + 1 )! 3 n + 1 = ( n + 1 )! 3 n + 1 ⋅ 3 n n !
Simplifying the Ratio We can simplify this expression: 3 n 3 n + 1 = 3 ( n + 1 )! n ! = ( n + 1 ) n ! n ! = n + 1 1 So, the ratio becomes: a n a n + 1 = n + 1 3
Computing the Limit Next, we compute the limit of this ratio as n approaches infinity: L = n → ∞ lim n + 1 3 As n becomes very large, the fraction n + 1 3 approaches 0. Therefore, L = 0
Applying the Ratio Test Since L = 0 < 1 , according to the Ratio Test, the series ∑ n = 1 ∞ n ! 3 n converges.
Conclusion Therefore, the series converges.
Examples
Consider a scenario in computer science where you're analyzing the efficiency of an algorithm. Suppose the number of operations the algorithm performs can be represented by the terms of a series. If this series converges, it means the algorithm's computational cost doesn't grow indefinitely as the input size increases, indicating an efficient algorithm. The Ratio Test helps determine if the series converges, providing insights into the algorithm's scalability and performance.
Using the Ratio Test on the series ∑ n = 1 ∞ n ! 3 n , we find that the limit of the ratio of consecutive terms is 0, which is less than 1. Thus, the series converges. This indicates that the terms of the series decrease in magnitude quickly enough to ensure the total sum remains finite.
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