1-decompose-the-fraction-F-x-1-x-3-x-1-4-2-find-the-sumA-n-1-1-n-3-n-1-4-and-B-n-1-1-n-n-3-n-1-4-3-what-is-the-value-of-n-0-1-n-1-

Question Number 104895 by mathmax by abdo last updated on 24/Jul/20

1) decompose the fraction F (x) = \frac{1}{x^{3} {(x + 1)}^{4}}

2) find the sumA = \sum_{n = 1}^{\infty} \frac{1}{n^{3} {(n + 1)}^{4}} and B = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{3} {(n + 1)}^{4}}

3) what is the value of \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{4} {(2 n + 1)}^{3}} ?

Answered by Ar Brandon last updated on 25/Jul/20

1 ∖ F (x) = \frac{1}{x^{3} {(x + 1)}^{4}} = \frac{a}{x + 1} + \frac{b}{{(x + 1)}^{2}} + \frac{c}{{(x + 1)}^{3}} + \frac{d}{{(x + 1)}^{4}} + \frac{{ex}^{2} + fx + g}{x^{3}}

F (x) = \frac{{ax}^{3} {(x + 1)}^{3} + {bx}^{3} {(x + 1)}^{2} + {cx}^{3} (x + 1) + {dx}^{3} + ({ex}^{2} + fx + g) {(x + 1)}^{4}}{x^{3} {(x + 1)}^{4}}

x \to - 1 \Rightarrow d = - 1, x \to 0 \Rightarrow g = 1, a + e \underset{x^{6}}{=} 0, 3 a + b + 4 e + f \underset{x^{5}}{=} 0,

3 a + 2 b + c + 6 e + 4 f + g \underset{x^{4}}{=} 0, a + b + c + d + 4 e + 6 f + 4 g \underset{x^{3}}{=} 0,

e + 4 f + 6 g \underset{x^{2}}{=} 0, f + 4 g \underset{x}{=} 0 \Rightarrow f = - 4, e = 10, a = - 10, b = - 6, c = - 3

\Rightarrow F (x) = \frac{- 10}{x + 1} + \frac{- 6}{{(x + 1)}^{2}} + \frac{- 3}{{(x + 1)}^{3}} + \frac{- 1}{{(x + 1)}^{4}} + \frac{{10 x}^{2} - 4 x + 1}{x^{3}}

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

1) F (x) = \frac{1}{x^{3} {(x + 1)}^{4}} \Rightarrow F (x) = \sum_{i = 1}^{3} \frac{a_{i}}{x^{i}} + \sum_{i = 1}^{4} \frac{b_{i}}{{(x + 1)}^{i}}

to find a_{i} we determine D_{2} (0) for f (x) = \frac{1}{{(x + 1)}^{4}} = {(x + 1)}^{- 4}

f (x) = f (0) + \frac{x}{1!} f^{^{'}} (0) + \frac{x^{2}}{2!} f^{(2)} (0) + \frac{x^{3}}{3!} ξ (x)

f (0) = 1, f^{^{'}} (x) = - 4 {(x + 1)}^{- 5} \Rightarrow f^{^{'}} (0) = - 4

f^{(2)} (x) = 20 {(x + 1)}^{- 6} \Rightarrow f^{(2)} (0) = 20 \Rightarrow f (x) = 1 - 4 x + {10 x}^{2} + \frac{x^{3}}{6} ξ (x) \Rightarrow

\frac{f (x)}{x^{3}} = \frac{1}{x^{3}} - \frac{4}{x^{2}} + \frac{10}{x} + \frac{1}{6} ξ (x) \Rightarrow a_{1} = 10, a_{2} = - 4, a_{3} = 1

to find b_{i} we determine D_{3} (- 1) for g (x) = \frac{1}{x^{3}} = x^{- 3} \Rightarrow

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

g (- 1) = - 1, g^{^{'}} (x) = - {3 x}^{- 4} \Rightarrow g^{^{'}} (- 1) = - 3

g^{(2)} (x) = 12 x^{- 5} \Rightarrow g^{(2)} (- 1) = - 12, g^{(3)} (x) = - 60 x^{- 6} \Rightarrow g^{(3)} (- 1) = - 60

\Rightarrow g (x) = - 1 - 3 (x + 1) - 6 {(x + 1)}^{2} - 10 {(x + 1)}^{3} + \frac{{(x + 1)}^{3}}{3!} ξ (x) \Rightarrow

\frac{g (x)}{{(x + 1)}^{4}} = - \frac{1}{{(x + 1)}^{4}} - \frac{3}{{(x + 1)}^{3}} - \frac{6}{{(x + 1)}^{2}} - \frac{10}{x + 1} + \frac{1}{4!} ξ (x) \Rightarrow

b_{1} = - 10, b_{2} = - 6, b_{3} = - 3, b_{4} = - 1 \Rightarrow

F (x) = \frac{10}{x} - \frac{4}{x^{2}} + \frac{1}{x^{3}} - \frac{10}{x + 1} - \frac{6}{{(x + 1)}^{2}} - \frac{3}{{(x + 1)}^{3}} - \frac{1}{{(x + 1)}^{4}}

Commented by mathmax by abdo last updated on 25/Jul/20

2) S = \sum_{n = 1}^{\infty} \frac{1}{n^{3} {(n + 1)}^{4}} \Rightarrow S = \lim_{n \to + \infty} \sum_{k = 1}^{n} \frac{1}{k^{3} {(k + 1)}^{4}}

we have \sum_{k = 1}^{n} \frac{1}{k^{3} {(k + 1)}^{4}} = 10 \sum_{k = 1}^{n} \frac{1}{k} - 4 \sum_{k = 1}^{n} \frac{1}{k^{2}} + \sum_{k = 1}^{n} \frac{1}{k^{3}} - 10 \sum_{k = 1}^{n} \frac{1}{k + 1}

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

10 \sum_{k = 1}^{n} (\frac{1}{k} - \frac{1}{k + 1}) = 10 (1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \dots . + \frac{1}{n} - \frac{1}{n + 1}) = 10 (1 - \frac{1}{n + 1}) \to 10

\sum_{k = 1}^{n} \frac{1}{k^{2}} \to \frac{π^{2}}{6} and \sum_{k = 1}^{n} \frac{1}{k^{3}} \to ξ (3)

\sum_{k = 1}^{n} \frac{1}{{(k + 1)}^{2}} = \sum_{k = 2}^{n + 1} \frac{1}{k^{2}} \to \frac{π^{2}}{6} - 1 and \sum_{k = 1}^{n} \frac{1}{{(k + 1)}^{3}} = \sum_{k = 2}^{n + 1} \frac{1}{k^{3}} \to ξ (3) - 1

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

S = 10 - 4 \times \frac{π^{2}}{6} + ξ (3) - 6 (\frac{π^{2}}{6} - 1) - 3 (ξ (3) - 1) - (ξ (4) - 1) \Rightarrow

S = 10 - \frac{2 π^{2}}{3} + ξ (3) - π^{2} + 6 - 3 ξ (3) + 3 - ξ (4) + 1

= 20 - \frac{5 π^{2}}{3} - 2 ξ (3) - ξ (4)