calculate-n-2-2n-1-n-4-n-2-

Question Number 45240 by maxmathsup by imad last updated on 10/Oct/18

c a l c u l a t e \sum_{n = 2}^{\infty} \frac{2 n + 1}{n^{4} - n^{2}}

Commented by maxmathsup by imad last updated on 11/Oct/18

l e t d e c o m p o s e F (x) = \frac{2 x + 1}{x^{4} - x^{2}} = \frac{2 x + 1}{x^{2} (x - 1) (x + 1)} \Rightarrow

F (x) = \frac{a}{x} + \frac{b}{x^{2}} + \frac{c}{x - 1} + \frac{d}{x + 1}

b = {l i m}_{x \to 0} x^{2} F (x) = - 1

c = {l i m}_{x \to 1} (x - 1) F (x) = \frac{3}{2}

d = {l i m}_{x \to - 1} (x + 1) F (x) = \frac{- 1}{- 2} = \frac{1}{2} \Rightarrow F (x) = \frac{a}{x} - \frac{1}{x^{2}} + \frac{3}{2 (x - 1)} + \frac{1}{2 (x + 1)}

F (2) = \frac{5}{4.3} = \frac{5}{12} = \frac{a}{2} - \frac{1}{4} + \frac{3}{2} + \frac{1}{6} \Rightarrow 5 = 6 a - 3 + 18 + 2 \Rightarrow 5 = 6 a + 17 \Rightarrow

6 a = - 12 \Rightarrow a = - 2 \Rightarrow F (x) = - \frac{2}{x} - \frac{1}{x^{2}} + \frac{3}{2 (x - 1)} + \frac{1}{2 (x + 1)}

l e t S_{n} = \sum_{k = 2}^{\infty} \frac{2 k + 1}{k^{4} - k^{2}} \Rightarrow S_{n} = \sum_{k = 2}^{n} F (k)

= - 2 \sum_{k = 2}^{n} \frac{1}{k} - \sum_{k = 2}^{n} \frac{1}{k^{2}} + \frac{3}{2} \sum_{k = 2}^{n} \frac{1}{k - 1} + \frac{1}{2} \sum_{k = 2}^{n} \frac{1}{k + 1} b u t

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

\sum_{k = 2}^{n} = ξ_{n} (2) - 1

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

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

S_{n} = - 2 H_{n} + 3 + ξ_{n} (2) - 1 + \frac{3}{2} H_{n - 1} + \frac{1}{2} H_{n + 1} - \frac{11}{12}

= - 2 H_{n} + \frac{3}{2} H_{n - 1} + \frac{1}{2} H_{n + 1} + ξ_{n} (2) + 2 - \frac{11}{12} \Rightarrow

S_{n} = - 2 (l n (n) + γ + o (\frac{1}{n})) + \frac{3}{2} (l n (n - 1) + γ + o (\frac{1}{n})) + \frac{1}{2} (l n (n) + γ + o (\frac{1}{n})) + ξ_{n} (2) + \frac{13}{12}

= - 2 l n (n) + \frac{3}{2} l n (n - 1) + \frac{1}{2} l n (n) + o (\frac{1}{n}) + ξ_{n} (2) + \frac{13}{12}

= l n (\sqrt{{(n - 1)}^{3}} \sqrt{n}) - l n (n^{2}) + o (\frac{1}{n}) + ξ_{n} (2) + \frac{13}{12}

= l n {\frac{(n - 1) \sqrt{n (n - 1)}}{n^{2}}} + o (\frac{1}{n}) + ξ_{n} (2) + \frac{13}{12} \Rightarrow

{l i m}_{n \to + \infty} S_{n} = ξ (2) + \frac{13}{12} = \frac{π^{2}}{6} + \frac{13}{12} .

Answered by ajfour last updated on 11/Oct/18

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

= Σ \frac{(n + 1) + (n - 1)}{(n - 1) n n (n + 1)} + \frac{1}{2} Σ \frac{(n + 1) - (n - 1)}{(n - 1) n n (n + 1)}

= Σ \frac{1}{(n - 1) n^{2}} + Σ \frac{1}{n^{2} (n + 1)}

+ \frac{1}{2} Σ \frac{1}{(n - 1) n^{2}} - \frac{1}{2} Σ \frac{1}{n^{2} (n + 1)}

= \frac{3}{2} Σ \frac{1}{(n - 1) n^{2}} + \frac{1}{2} Σ \frac{1}{n^{2} (n + 1)}

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

= \frac{3}{4} Σ (\frac{1}{n - 1} - \frac{1}{n}) - \frac{1}{4} Σ (\frac{1}{n} - \frac{1}{n + 1})

- \frac{1}{2} Σ \frac{1}{n^{2}}

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

- \frac{1}{2} (\frac{1}{4} + \frac{1}{9} + \dots)

= \frac{3}{4} - \frac{1}{8} - \frac{1}{2} (\frac{1}{4} + \frac{1}{9} + \frac{1}{16} + . .)

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

\int_{0}^{1} \frac{\ln (1 - x)}{x} d x = - (1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + . .)

s o

\underset{n = 2}{\sum^{\infty}} \frac{2 n + 1}{n^{4} - n^{2}} = \frac{3}{4} - \frac{1}{8} - \frac{1}{2} [1 + \int_{0}^{1} \frac{\ln (1 - x)}{x} d x]

\dots . . (c a n t s o l v e t h e i n t e g r a l, s i r) .

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

\frac{2 n + 1}{n^{2} (n + 1) (n - 1)} = \frac{a}{n} + \frac{b}{n^{2}} + \frac{c}{n + 1} + \frac{d}{n - 1}

2 n + 1 = a n (n + 1) (n - 1) + b (n + 1) (n - 1) + {c n}^{2} (n - 1) + {d n}^{2} (n + 1)

p u t n = 0

1 = b (- 1) b = - 1

p u t n - 1 = 0

3 = d \times 1 \times 2 d = \frac{3}{2}

p u t n + 1 = 0

- 2 + 1 = c \times {(- 1)}^{2} \times (- 2)

- 1 = - 2 c c = \frac{1}{2}

2 n + 1 = a n (n + 1) (n - 1) + b (n + 1) (n - 1) + {c n}^{2} (n - 1) + {d n}^{2} (n + 1)

2 n + 1 = a (n^{3} - n) + (- 1) (n^{2} - 1) + \frac{1}{2} n^{2} (n - 1) + \frac{3}{2} n^{2} (n + 1)

2 n + 1 = n^{3} (a + \frac{1}{2} + \frac{3}{2}) + n^{2} (- 1 - \frac{1}{2} + \frac{3}{2}) + n (- a) + 1

a + 2 = 0 a = - 2

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

= S_{1} + S_{2} + S_{3} + S_{4}

S_{1} = - 2 \underset{n = 2}{\sum^{\infty}} \frac{1}{n}

S_{1} = - 2 {(\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \dots \infty) - \frac{1}{1}}

= - 2 (γ - 1) = - 2 γ + 1 γ = E u l e r s c o n s t a n t

S_{2} = - 1 \underset{n = 2}{\sum^{\infty}} \frac{1}{n^{2}}

= - 1 {(\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . \infty) - \frac{1}{1}}

= - (\frac{π^{2}}{6} - 1) = 1 - \frac{π^{2}}{6}

S_{3} = \frac{1}{2} \underset{n = 2}{\sum^{\infty}} \frac{1}{n + 1}

= \frac{1}{2} {(\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \dots \infty) - (\frac{1}{1} + \frac{1}{2})}

= \frac{1}{2} (γ - \frac{3}{2}) = \frac{γ}{2} - \frac{3}{4}

S_{4} = \frac{3}{2} \underset{n = 2}{\sum^{\infty}} \frac{1}{n - 1}

\frac{3}{2} (\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \dots)

= \frac{3 γ}{2}

s o r e a u i r e d a n s i s S_{1} + S_{2} + S_{3} + S_{4}

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

= - 2 γ + 2 γ + 1 + 1 - \frac{3}{4} - \frac{π^{2}}{6}

= \frac{5}{4} - \frac{π^{2}}{6} p l s c h e c k \dots