calculate-k-0-n-1-k-k-1-3-C-n-k-

Question Number 100237 by mathmax by abdo last updated on 25/Jun/20

calculate \sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{{(k + 1)}^{3}} C_{n}^{k}

Answered by maths mind last updated on 26/Jun/20

x^{a} {(1 - x)}^{n} = Σ x^{a} C_{n}^{k} {(- x)}^{k} = Σ C_{n}^{k} {(- 1)}^{k} x^{a + k}

\Rightarrow \int_{0}^{1} x^{a} {(1 - x)}^{n} d x = \underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k} C_{n}^{k}}{a + k + 1} \dots E

β (a, b) = \int_{0}^{1} x^{a - 1} {(1 - x)}^{b - 1} d x

E = β (a + 1, n + 1) = \underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k} C_{n}^{k}}{a + k + 1}

\Rightarrow \frac{\partial^{2}}{\partial a^{2}} β (a + 1, n + 1) = \underset{k = 0}{\sum^{n}} \frac{2 {(- 1)}^{k} C_{n}^{k}}{{(a + k + 1)}^{3}}

\Leftrightarrow

\frac{\partial^{2}}{\partial a^{2}} β (a + 1, n + 1) ∣_{a = 0} = \underset{k = 0}{\sum^{n}} \frac{2 {(- 1)}^{k} C_{n}^{k}}{{(k + 1)}^{3}}

\frac{\partial β (a + 1, n + 1)}{\partial a} = β (a + 1, n + 1) (Ψ (a + 1) - Ψ (n + a + 2))

\Rightarrow β (a + 1, n + 1) {(Ψ (a + 1) - Ψ (n + a + 2))}^{2} + β (a + 1, n + 1) (Ψ^{'} (a + 1) - Ψ^{'} (n + a + 2))

\Leftrightarrow

Σ 2 \frac{{(- 1)}^{k}}{{(k + 1)}^{3}} C_{n}^{k} = β (1, n + 1) (Ψ (1) - Ψ (n + 2)) + β (1, n + 1) (Ψ^{'} (1) - Ψ^{'} (n + 2))

β (1, n + 1) = \frac{1}{n + 1}

Ψ (1) - Ψ (n + 2) = - H_{n + 1}

Ψ^{'} (z) = ζ (2, z) = \underset{k = 0}{\sum^{\infty}} \frac{1}{{(z + k)}^{2}}

Ψ^{'} (1) - Ψ^{'} (n + 2) = - H_{n + 1}^{(2)}

w e g e t

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

\Rightarrow - \frac{1}{2 (n + 1)} (H_{n + 1} + H_{n + 1}^{(2)}) = \underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k} C_{n}^{k}}{{(k + 1)}^{3}}

Commented by mathmax by abdo last updated on 26/Jun/20

thank you sir .