$\sum_{n=1}^{\infty} \frac{1}{4 n^2-25}$
Answer (2)
Perform partial fraction decomposition: Express 4 n 2 − 25 1 as 10 1 ( 2 n − 5 1 − 2 n + 5 1 ) .
Rewrite the sum: Express the original sum as 10 1 ∑ n = 1 ∞ ( 2 n − 5 1 − 2 n + 5 1 ) .
Identify the telescoping pattern: Observe that the series telescopes, with terms canceling out.
Evaluate the limit: Determine the limit of the partial sums as n approaches infinity, which is 50 1 .
50 1
Explanation
Problem Analysis We are asked to find the sum of the infinite series ∑ n = 1 ∞ 4 n 2 − 25 1 . This problem involves evaluating an infinite sum, which can often be solved using partial fraction decomposition and telescoping series.
Partial Fraction Decomposition First, we perform partial fraction decomposition on the summand. We can factor the denominator as 4 n 2 − 25 = ( 2 n − 5 ) ( 2 n + 5 ) . Thus, we want to find constants A and B such that 4 n 2 − 25 1 = 2 n − 5 A + 2 n + 5 B .
Solving for A and B Multiplying both sides by ( 2 n − 5 ) ( 2 n + 5 ) , we get 1 = A ( 2 n + 5 ) + B ( 2 n − 5 ) . To solve for A and B , we can use convenient values of n . If n = 2 5 , then 1 = A ( 2 ( 2 5 ) + 5 ) + B ( 0 ) , so 1 = 10 A , which means A = 10 1 . If n = − 2 5 , then 1 = A ( 0 ) + B ( 2 ( − 2 5 ) − 5 ) , so 1 = − 10 B , which means B = − 10 1 . Therefore, 4 n 2 − 25 1 = 2 n − 5 1/10 − 2 n + 5 1/10 = 10 1 ( 2 n − 5 1 − 2 n + 5 1 ) .
Rewriting the Sum Now we can rewrite the sum as n = 1 ∑ ∞ 4 n 2 − 25 1 = 10 1 n = 1 ∑ ∞ ( 2 n − 5 1 − 2 n + 5 1 ) . Let's write out the first few terms of the series to see if we can identify a telescoping pattern: \begin{align*} \sum_{n=1}^{\infty} \left( \frac{1}{2n-5} - \frac{1}{2n+5} \right) &= \left( \frac{1}{-3} - \frac{1}{7} \right) + \left( \frac{1}{-1} - \frac{1}{9} \right) + \left( \frac{1}{1} - \frac{1}{11} \right) + \left( \frac{1}{3} - \frac{1}{13} \right) + \left( \frac{1}{5} - \frac{1}{15} \right) + \left( \frac{1}{7} - \frac{1}{17} \right) + \left( \frac{1}{9} - \frac{1}{19} \right) + \left( \frac{1}{11} - \frac{1}{21} \right) + \cdots \end{align*}
Telescoping Series We can see that some terms will cancel out. To find the sum, we consider the partial sums. Let S N be the N -th partial sum: S N = n = 1 ∑ N ( 2 n − 5 1 − 2 n + 5 1 ) . We can rewrite this as S N = n = 1 ∑ N 2 n − 5 1 − n = 1 ∑ N 2 n + 5 1 . Let's shift the index of the second sum by setting m = n + 5 , so n = m − 5 . When n = 1 , m = 6 , and when n = N , m = N + 5 . Thus, S N = n = 1 ∑ N 2 n − 5 1 − m = 6 ∑ N + 5 2 ( m − 5 ) + 5 1 = n = 1 ∑ N 2 n − 5 1 − n = 6 ∑ N + 5 2 n − 5 1 . Now we can write out the terms that don't cancel: S N = ( − 3 1 + − 1 1 + 1 1 + 3 1 + 5 1 ) − ( 2 N + 1 1 + 2 N + 3 1 + 2 N + 5 1 + 2 N − 1 1 + 2 N − 3 1 ) . S N = − 3 1 − 1 + 1 + 3 1 + 5 1 − 2 N + 1 1 − 2 N + 3 1 − 2 N + 5 1 − 2 N − 1 1 − 2 N − 3 1 = 5 1 − ( 2 N − 3 1 + 2 N − 1 1 + 2 N + 1 1 + 2 N + 3 1 + 2 N + 5 1 ) .
Evaluating the Limit As N → ∞ , the terms in the parenthesis approach 0. Thus, N → ∞ lim S N = 5 1 . Therefore, n = 1 ∑ ∞ 4 n 2 − 25 1 = 10 1 n = 1 ∑ ∞ ( 2 n − 5 1 − 2 n + 5 1 ) = 10 1 ⋅ 5 1 = 50 1 .
Final Answer Thus, the sum of the series is 50 1 .
Examples
Consider a scenario where you are analyzing the stability of a structure subjected to a series of oscillating forces. Each term in the series represents the amplitude of a particular harmonic component of the force. Evaluating the infinite sum allows you to determine the overall stability of the structure by understanding how these harmonic components interact. This is crucial in engineering to prevent resonance and ensure the structure's integrity.
The infinite series ∑ n = 1 ∞ 4 n 2 − 25 1 can be evaluated using partial fraction decomposition, which simplifies it to a telescoping series. After decomposing and rewriting, the sum converges to 50 1 . Thus, the final answer is 50 1 .
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