[tex]$\sum_{n=1}^{\infty} \frac{2^n(x-5)^n}{n^2}$[/tex]
Answer (1)
Apply the ratio test to the series ∑ n = 1 ∞ n 2 2 n ( x − 5 ) n to find the radius of convergence.
Determine the limit: lim n → ∞ a n a n + 1 = 2∣ x − 5∣ .
Set 2∣ x − 5∣ < 1 to find the interval ( 2 9 , 2 11 ) .
Check endpoints x = 2 9 and x = 2 11 and conclude the interval of convergence is [ 2 9 , 2 11 ] .
Explanation
Problem Setup We are given the power series ∑ n = 1 ∞ n 2 2 n ( x − 5 ) n and we want to find its interval of convergence. To do this, we will use the ratio test.
Applying the Ratio Test The ratio test states that if lim n → ∞ a n a n + 1 < 1 , then the series converges. In our case, a n = n 2 2 n ( x − 5 ) n . Let's compute the ratio a n a n + 1 :
Calculating the Ratio a n a n + 1 = n 2 2 n ( x − 5 ) n ( n + 1 ) 2 2 n + 1 ( x − 5 ) n + 1 = ( n + 1 ) 2 2 n + 1 ( x − 5 ) n + 1 ⋅ 2 n ( x − 5 ) n n 2
Simplifying the Ratio Now, we simplify the expression:
Further Simplification ( n + 1 ) 2 2 n + 1 ( x − 5 ) n + 1 ⋅ 2 n ( x − 5 ) n n 2 = 2 ( x − 5 ) ( n + 1 ) 2 n 2 = 2∣ x − 5∣ ( n + 1 n ) 2
Taking the Limit Next, we take the limit as n → ∞ :
Evaluating the Limit n → ∞ lim 2∣ x − 5∣ ( n + 1 n ) 2 = 2∣ x − 5∣ n → ∞ lim ( n + 1 n ) 2 = 2∣ x − 5∣ ⋅ 1 2 = 2∣ x − 5∣
Finding the Interval of Convergence For the series to converge, we need 2∣ x − 5∣ < 1 , which means ∣ x − 5∣ < 2 1 . This gives us the interval 5 − 2 1 < x < 5 + 2 1 , or 2 9 < x < 2 11 .
Checking the Endpoints Now, we need to check the endpoints of the interval, x = 2 9 and x = 2 11 .
Endpoint x = 2 9 If x = 2 9 , the series becomes:
Convergence at x = 2 9 n = 1 ∑ ∞ n 2 2 n ( 2 9 − 5 ) n = n = 1 ∑ ∞ n 2 2 n ( − 2 1 ) n = n = 1 ∑ ∞ n 2 ( − 1 ) n This is an alternating series with terms n 2 1 that decrease to 0. Since ∑ n = 1 ∞ n 2 1 converges (p-series with p=2 > 1), the alternating series also converges.
Endpoint x = 2 11 If x = 2 11 , the series becomes:
Convergence at x = 2 11 n = 1 ∑ ∞ n 2 2 n ( 2 11 − 5 ) n = n = 1 ∑ ∞ n 2 2 n ( 2 1 ) n = n = 1 ∑ ∞ n 2 1 This is a p-series with 1"> p = 2 > 1 , so it converges.
Final Interval of Convergence Since the series converges at both endpoints, the interval of convergence is [ 2 9 , 2 11 ] .
Conclusion Therefore, the interval of convergence for the given power series is [ 2 9 , 2 11 ] .
Examples
Power series are used extensively in engineering and physics to approximate functions, especially when dealing with complex systems where closed-form solutions are not available. For instance, in signal processing, a power series might represent a filter's transfer function, and determining its interval of convergence ensures the filter remains stable and predictable. Similarly, in thermodynamics, power series can model the behavior of materials under varying conditions, and the convergence interval dictates the range of validity for these models.
