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Question Number 182794 by mnjuly1970 last updated on 14/Dec/22

Ω = \underset{n = 1}{\sum^{\infty}} \frac{{(- \underset{}{\overset{}{1}})}^{n}}{2^{n} . n (n \underset{}{\overset{}{+}} 1) (n \underset{}{\overset{}{+}} 2)} = ?

Answered by ARUNG_Brandon_MBU last updated on 14/Dec/22

S (x) = \underset{n = 1}{\sum^{\infty}} \frac{x^{n + 2}}{n (n + 1) (n + 2)} \Rightarrow S^{'} (x) = \underset{n = 1}{\sum^{\infty}} \frac{x^{n + 1}}{n (n + 1)}

\Rightarrow S^{″} (x) = \underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n} \Rightarrow S^{‴} (x) = \underset{n = 1}{\sum^{\infty}} x^{n - 1} = \frac{1}{1 - x}

\Rightarrow S^{″} (x) = \underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n} = - \ln (1 - x) + C_{1}, C_{1} = 0

\Rightarrow S^{'} (x) = \underset{n = 1}{\sum^{\infty}} \frac{x^{n + 1}}{n (n + 1)} = (1 - x) \ln (1 - x) + x - 1 + C_{2}, C_{2} = 1

\Rightarrow S^{'} (x) = \ln (1 - x) - x \ln (1 - x) + x

\Rightarrow S (x) = \int [- x \ln (1 - x) + \ln (1 - x) + x] d x

\Rightarrow S (x) = - \frac{x^{2}}{2} \ln (1 - x) + \frac{1}{2} (\frac{x^{2}}{2} + x + \ln ∣ x - 1 ∣)

+ (1 - x) - (1 - x) \ln (1 - x) + \frac{x^{2}}{2} + C_{3}, C_{3} = - 1

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

= - \frac{1}{2} \ln (\frac{3}{2}) + 2 (\frac{1}{8} - \frac{1}{2} + \ln (\frac{3}{2})) + 6 - 6 \ln (\frac{3}{2}) + \frac{1}{2} - 4

\Rightarrow Ω = \frac{7}{4} - \frac{9}{2} \ln (\frac{3}{2})

Commented by mnjuly1970 last updated on 14/Dec/22

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