I would like to find the exact value of the series∑n=3∞4n−3n(n2−4)

4n−3n(n2−4)=34n−118(n+2)+58(n−2) 4n−3n(n2−4)=18(6(1n−1n+2)+5(1n−2−1n+2)) ∑n=3m1n−1n+2=712−1m+1−1m+2→72 ∑n=3m1n−2−1n+2=2512−1m−1−1m−1m+1−1m+2→2512 ∑n=3∞4n−3n(n2−4)=18(6×712+5×2512)=16796

Answered: "I would like to find the exact value..."

High School
Calculus and Analysis
Calculus 2
Series

iristh3virusoo2

Answered question

2022-02-23

I would like to find the exact value of the series

\sum_{n = 3}^{\infty} \frac{4 n - 3}{n (n^{2} - 4)}

Answer & Explanation

mljaskaj0u

Beginner2022-02-24Added 2 answers

\frac{4 n - 3}{n (n^{2} - 4)} = \frac{3}{4 n} - \frac{11}{8 (n + 2)} + \frac{5}{8 (n - 2)}

\frac{4 n - 3}{n (n^{2} - 4)} = \frac{1}{8} (6 (\frac{1}{n} - \frac{1}{n + 2}) + 5 (\frac{1}{n - 2} - \frac{1}{n + 2}))

\sum_{n = 3}^{m} \frac{1}{n} - \frac{1}{n + 2} = \frac{7}{12} - \frac{1}{m + 1} - \frac{1}{m + 2} \to \frac{7}{2}

\sum_{n = 3}^{m} \frac{1}{n - 2} - \frac{1}{n + 2} = \frac{25}{12} - \frac{1}{m - 1} - \frac{1}{m} - \frac{1}{m + 1} - \frac{1}{m + 2} \to \frac{25}{12}

\sum_{n = 3}^{\infty} \frac{4 n - 3}{n (n^{2} - 4)} = \frac{1}{8} (6 \times \frac{7}{12} + 5 \times \frac{25}{12}) = \frac{167}{96}

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