calculate-n-1-1-n-n-2-n-2-3-

Question Number 95217 by mathmax by abdo last updated on 24/May/20

calculate \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2} {(n + 2)}^{3}}

Answered by mathmax by abdo last updated on 25/May/20

let decompose F (x) = \frac{1}{x^{2} {(x + 2)}^{3}} \Rightarrow F (x) = \sum_{i = 1}^{2} \frac{a_{i}}{x^{i}} + \sum_{i = 1}^{3} \frac{b_{i}}{{(x + 2)}^{i}}

a_{i} ? we find D_{1} (0) for f (x) = {(x + 2)}^{- 3} \Rightarrow f^{^{'}} (x) = - 3 {(x + 2)}^{- 4}

f (x) = f (o) + {xf}^{^{'}} (0) + \frac{x^{2}}{2!} ξ (x) = 2^{- 3} - 3 . 2^{- 4} x + \frac{x^{2}}{2} ξ (x) \Rightarrow

\frac{f (x)}{x^{2}} = \frac{2^{- 3}}{x^{2}} - \frac{3 . 2^{- 4}}{x} + \frac{x}{2} ξ (x) \Rightarrow a_{1} = - \frac{3}{2^{4}} and a_{2} = \frac{1}{2^{3}}

b_{i} we find D_{2} (- 2) for g (x) = x^{- 2} \Rightarrow g^{^{'}} (x) = - {2 x}^{- 3} \Rightarrow g^{(2)} (x) = {6 x}^{- 4}

g (x) = g (- 2) + \frac{x + 2}{!} g^{^{'}} (- 2) + \frac{{(x + 2)}^{2}}{2!} g^{(2)} (- 2) + \frac{{(x + 2)}^{3}}{3!} ξ (x)

= {(- 2)}^{- 2} + (x + 2) (- 2) {(- 2)}^{- 3} + \frac{{(x + 2)}^{2}}{2} (6) {(- 2)}^{- 4} + \frac{{(x + 2)}^{3}}{6} ξ (x)

\frac{g (x)}{{(x + 2)}^{3}} = \frac{{(- 2)}^{- 2}}{{(x + 2)}^{3}} + \frac{{(- 2)}^{- 2}}{{(x + 2)}^{2}} + \frac{3 {(- 2)}^{- 4}}{x + 2} + \frac{1}{6} ξ (x) \Rightarrow

b_{1} = \frac{3}{16}, b_{2} = \frac{1}{4}, b_{3} = \frac{1}{4} \Rightarrow

F (x) = - \frac{3}{16 x} + \frac{1}{{8 x}^{2}} + \frac{3}{16 (x + 2)} + \frac{1}{4 {(x + 2)}^{2}} + \frac{1}{4 {(x + 2)}^{3}} \Rightarrow

\sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k^{2} {(k + 1)}^{3}} = - \frac{3}{16} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k} + \frac{1}{8} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k^{2}} + \frac{3}{16} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k + 2}

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

\sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k} \to - \ln (2) (n \to + \infty)

\sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k^{2}} \to δ (2) = (2^{1 - 2} - 1) ξ (2) = - \frac{1}{2} \times \frac{π^{2}}{6} = - \frac{π^{2}}{12}

\sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k + 2} = \sum_{k = 3}^{n + 2} \frac{{(- 1)}^{k}}{k} \to \sum_{k = 3}^{\infty} \frac{{(- 1)}^{k}}{k}

= - \ln 2 - (- 1 + \frac{1}{2}) = - \ln 2 + \frac{1}{2}

\sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{{(k + 2)}^{2}} = \sum_{k = 3}^{n + 2} \frac{{(- 1)}^{k}}{k^{2}} \to \sum_{k = 3}^{\infty} \frac{{(- 1)}^{k}}{k^{2}} = - \frac{π^{2}}{12} - (- 1 + \frac{1}{4})

= \frac{3}{4} - \frac{π^{2}}{12}

\sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{{(k + 2)}^{3}} = \sum_{k = 3}^{n + 2} \frac{{(- 1)}^{k}}{k^{3}} \to \sum_{k = 3}^{\infty} \frac{{(- 1)}^{k}}{k^{3}} - (- 1 + \frac{1}{8})

= (2^{1 - 3} - 1) ξ (3) + \frac{7}{8} = (\frac{1}{4} - 1) ξ (3) + \frac{7}{8} = \frac{7}{8} - \frac{3}{4} ξ (3)

the value of this sum is known \dots