n-1-1-n-2n-1-2-

Question Number 160289 by amin96 last updated on 27/Nov/21

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

Answered by qaz last updated on 27/Nov/21

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

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

= H_{1 / 2} - 2 ((1 - 2^{- 2}) ζ (2) - 1)

= 4 - 2 \ln 2 - \frac{3}{2} ζ (2)

Answered by mathmax by abdo last updated on 28/Nov/21

we decompose f (x) = \frac{1}{x {(2 x + 1)}^{2}} \Rightarrow f (x) = \frac{a}{x} + \frac{b}{2 x + 1} + \frac{c}{{(2 x + 1)}^{2}}

a = xf (x) ∣_{x = 0} = 1

c = {(2 x + 1)}^{2} f (x) ∣_{x = - \frac{1}{2}} = - 2

\lim_{x \to + \infty} xf (x) = 0 = a + \frac{b}{2} \Rightarrow \frac{b}{2} = - a \Rightarrow b = - 2 \Rightarrow

f (x) = \frac{1}{x} - \frac{2}{2 x + 1} - \frac{2}{{(2 x + 1)}^{2}}

S_{n} = \sum_{k = 1}^{n} \frac{1}{k {(2 k + 1)}^{2}} \Rightarrow S_{n} = \sum_{k = 1}^{n} f (k) = \sum_{k = 1}^{n} (\frac{1}{k} - \frac{2}{2 k + 1} - \frac{2}{{(2 k + 1)}^{2}})

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

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

\sum_{k = 1}^{n} \frac{1}{2 k + 1} = \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{2 n + 1} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots . + \frac{1}{2 n} + \frac{1}{2 n + 1}

- 1 - \frac{1}{2} - \frac{1}{4} - \dots - \frac{1}{2 n} = H_{2 n + 1} - 1 - \frac{1}{2} H_{n}

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

Σ \frac{1}{k^{2}} = \frac{1}{4} Σ \frac{1}{k^{2}} + \sum_{k = 0}^{\infty} \frac{1}{{(2 k + 1)}^{2}} \Rightarrow \sum_{k = 0}^{\infty} \frac{1}{{(2 k + 1)}^{2}} = \frac{3}{4} \frac{π^{2}}{6} = \frac{π^{2}}{8}

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

= {2 H}_{n} - {2 H}_{2 n + 1} + 2 \sim 2 (\ln (n) + γ) - 2 (\ln (2 n + 1) + γ) + 2

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

\lim_{n \to + \infty} S_{n} = 2 - 2 \ln 2 - 2 {\frac{π^{2}}{8} - 1}

= 2 - 2 \ln 2 - \frac{π^{2}}{4} + 2 = 4 - 2 \ln 2 - \frac{π^{2}}{4}