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Question Number 147572 by liberty last updated on 22/Jul/21

\sum_{n ⩾ 1} \frac{4 n - 3}{(n^{2} + 2 n) (n + 3)} = ?

Answered by mathmax by abdo last updated on 22/Jul/21

let S = \sum_{n = 1}^{\infty} \frac{4 n - 3}{n (n + 2) (n + 3)} let decompose F (x) = \frac{4 x - 3}{x (x + 2) (x + 3)}

F (x) = \frac{a}{x} + \frac{b}{x + 2} + \frac{c}{x + 3}

a = \frac{- 3}{6} = - \frac{1}{2}, b = \frac{- 11}{(- 2) (1)} = \frac{11}{2}, c = \frac{- 15}{(- 3) (- 1)} = - 5 \Rightarrow

F (x) = - \frac{1}{2 x} + \frac{11}{2 (x + 2)} - \frac{5}{x + 3} \Rightarrow

\sum_{k = 1}^{n} F (k) = - \frac{1}{2} \sum_{k = 1}^{n} \frac{1}{k} + \frac{11}{2} \sum_{k = 1}^{n} \frac{1}{k + 2} - 5 \sum_{k = 1}^{n} \frac{1}{k + 3} = S_{n}

\sum_{k = 1}^{n} \frac{1}{k} = H_{n}, \sum_{k = 1}^{n} \frac{1}{k + 2} = \sum_{k = 3}^{n + 2} \frac{1}{k} = H_{n + 2} - \frac{3}{2}

\sum_{k = 1}^{n} \frac{1}{k + 3} = \sum_{k = 1}^{n + 4} \frac{1}{k} = H_{n + 4} - \frac{3}{2} - \frac{1}{3} = H_{n + 4} - \frac{11}{6} \Rightarrow

S_{n} = - \frac{1}{2} H_{n} + \frac{11}{2} (H_{n + 2} - \frac{3}{2}) - 5 (H_{n + 4} - \frac{11}{6})

= - \frac{1}{2} H_{n} + \frac{11}{2} H_{n + 2} - \frac{33}{4} + {5 H}_{n + 4} - \frac{55}{6}

S_{n} = - \frac{1}{2} (logn + γ + o (\frac{1}{n}) + \frac{11}{2} (\log (n + 2) + γ + o (\frac{1}{n + 2}))

- 5 (\log (n + 4) + γ + o (\frac{1}{n + 4})) - \frac{33}{4} + \frac{55}{6} \Rightarrow

S_{n} \sim - \frac{1}{2} logn - \frac{γ}{2} + \frac{11}{2} logn + \frac{11}{2} \log (1 + \frac{2}{n}) + \frac{11 γ}{2} - 5 \log (n) - 5 \log (1 + \frac{4}{n})

- 5 γ - \frac{33}{4} + \frac{55}{6} \Rightarrow S_{n} \sim \frac{11}{2} \log (1 + \frac{2}{n}) - 5 \log (1 + \frac{4}{n}) - \frac{33}{4} + \frac{55}{6} \Rightarrow

\lim_{n \to \infty} S_{n} = \frac{55}{6} - \frac{33}{4}

Answered by qaz last updated on 22/Jul/21

\underset{n = 1}{\sum^{\infty}} \frac{4 n - 3}{n (n + 2) (n + 3)}

= \underset{n = 1}{\sum^{\infty}} (- \frac{1}{2 n} + \frac{11}{2} \cdot \frac{1}{n + 2} - 5 \cdot \frac{1}{n + 3})

= \underset{n = 0}{\sum^{\infty}} (- \frac{1}{2} \cdot \frac{1}{n + 1} + \frac{11}{2} \cdot \frac{1}{n + 3} - 5 \cdot \frac{1}{n + 4})

= \underset{n = 0}{\sum^{\infty}} (- \frac{1}{2} \int_{0}^{1} x^{n} dx + \frac{11}{2} \int_{0}^{1} x^{n + 2} dx - 5 \int_{0}^{1} x^{n + 3} dx)

= - \frac{1}{2} \int_{0}^{1} \frac{dx}{1 - x} dx + \frac{11}{2} \int_{0}^{1} \frac{x^{2}}{1 - x} dx - 5 \int_{0}^{1} \frac{x^{3}}{1 - x} dx

= \frac{1}{2} \int_{0}^{1} \frac{- 1 + {11 x}^{2} - {10 x}^{3}}{1 - x} dx

= - \frac{1}{2} \int_{0}^{1} (1 + x - {10 x}^{2}) dx

= \frac{11}{12}

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