find-n-1-1-n-n-2-n-1-3-n-2-4-

Question Number 116555 by Bird last updated on 04/Oct/20

f i n d \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2} {(n + 1)}^{3} {(n + 2)}^{4}}

Answered by Olaf last updated on 05/Oct/20

R (n) = \frac{1}{n^{2} {(n + 1)}^{3} {(n + 2)}^{4}}

R (n) = - \frac{5}{16} . \frac{1}{n} + \frac{1}{16} . \frac{1}{n^{2}} + \frac{5}{n + 1} - \frac{2}{{(n + 1)}^{2}}

+ \frac{1}{{(n + 1)}^{3}} - \frac{75}{16} . \frac{1}{n + 2} - \frac{39}{16} . \frac{1}{{(n + 2)}^{2}}

- \frac{1}{{(n + 2)}^{3}} - \frac{1}{4} . \frac{1}{{(n + 2)}^{4}}

\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n} = - \ln 2

\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2}} = - \frac{π^{2}}{12}

\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{3}} = - \frac{3}{12} ξ (3)

\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{4}} = - \frac{7 π^{4}}{720}

S = - \frac{5}{16} (- \ln 2) + \frac{1}{16} (- \frac{π^{2}}{12}) + 5 (- \ln 2 + 1)

- 2 (- \frac{π^{2}}{12} + 1) + (- \frac{3}{12} ξ (3) + 1) - \frac{75}{16} (- \ln 2 + 1 - \frac{1}{2})

- \frac{39}{16} (- \frac{π^{2}}{12} + 1 - \frac{1}{4}) - (- \frac{3}{12} ξ (3) + 1 - \frac{1}{8})

- \frac{1}{4} (- \frac{7 π^{4}}{720} + 1 - \frac{1}{16})

S = \ln 2 (\frac{5}{16} - 5 + \frac{75}{16})

- \frac{π^{2}}{12} (\frac{1}{16} - 2 - \frac{39}{16})

- \frac{3}{12} ξ (3) (1 - 1) + \frac{7 π^{4}}{2880}

+ 5 - 2 + 1 - \frac{75}{32} - \frac{117}{64} - \frac{7}{8} - \frac{15}{64}

S = \frac{35 π^{2}}{96} + \frac{7 π^{4}}{2880} - \frac{41}{32}

Please mister verify the calculous .

Commented by mathmax by abdo last updated on 05/Oct/20

thank you sir .