Category Archives: Striling numbers of first kind

$$\int\limits_0^1 \dfrac{\log^{m} (1+x)\log^n x}{x}\; dx = (-1)^{n+1}(n!) (m)! \sum_{\{m-1\}} \sum_{k=1}^\infty \frac{(-1)^k}{k^{n+2}} \prod^{l’}_{j=1}\frac{(-1)^{i_j}}{(i_j)!} \left( \frac{H_{k-1}^{(r_j)}}{r_j}\right)^{i_j}$$ $$\textit{solution}$$ Stirling numbers of the first kind might be useful here, Consider $$m! \sum_{k=m}^\infty (-1)^{k-m} \left[k\atop m\right] \frac{x^k}{k!} = \log^m(1+x)$$ $$\int\limits_0^1 \dfrac{\log^m (1+x)\log^n x}{x}\; dx = … Continue reading | | Leave a comment Nonlinear euler sum proof using stirling numbers of the first kind Prove that$$\sum_{k=1}^\infty \frac{(H_k)^2}{k^2} = \frac{17\pi^4}{360}\textit{proof}$$Start by the following which can be proved by induction$$\frac{\left[n\atop 3\right]}{n!} =\frac{ (H_{n-1})^2-H^{(2)}_{n-1}}{2n}$$And the generating function proved here$$-\sum_{n=3}^\infty \left[n\atop 3\right] \frac{z^n}{n!} = \frac{\log^3(1-z)}{6}$$Hence we get$$\sum_{n=3}^\infty ( H^{(2)}_{n-1}- (H_{n-1})^2) … Continue reading

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Stirling numbers of first kind generating function

Prove the following $$\sum_{n=k}^\infty(-1)^{n-k}\left[n\atop k\right] \frac{z^n}{n!} = \frac{\log^k(1+z)}{k!}$$ $$\textit{proof}$$ We start by the following $$(1+z)^u = \sum_{n=0}^\infty {u \choose n} z^n$$ Now use that $${u \choose n} = \frac{\Gamma(u+1)}{\Gamma(u-n+1)n!}$$ Now use that $$\frac{\Gamma(u+1)}{\Gamma(u-n+1)} = \frac{u(u-1)\cdots (u-n+1)\Gamma(u+1)}{\Gamma(u+1)} = (u)_n$$ This implies … Continue reading

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Signed Stirling numbers of first kind as coefficients

Signed Stirling numbers of the first kind We define the following $$s(n,k) = (-1)^{n-k} \left[n\atop k\right]$$ Prove the following $$(x)_n = x(x-1)(x-2)\cdots (x-n+1) = \sum_{k=0}^n s(n,k)x^k$$ $$\textit{proof}$$ We already proved that $$x^{(n)} = \sum_{k=0}^n \left[n\atop k\right] x^k$$ Which can be … Continue reading

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Prove that $$x^{(n)} = x(x+1)(x+2)\cdots(x+n-1) = \sum_{k=0}^n \left[n\atop k\right]x^k$$ Or $$\sum_{k=0}^n \left[n\atop k\right]x^k = \frac{\Gamma(x+n)}{\Gamma(x)}$$ $$\textit{proof}$$ By induction for $$n=0$$ we have $$\left[0\atop 0\right] = 1$$ Assume it is true for $$n$$ then for $$… Continue reading Posted in Striling numbers of first kind | | Leave a comment Relation between harmonic numbers and Stirling numbers of the first kind Prove that \left[n\atop 2\right] = H_{n-1}\Gamma(n) By induction on \( n$$ we have for $$n=2$$ $$\left[2\atop 2\right] = H_{1}\times\Gamma(1) = 1$$ Assume that $$\left[k\atop 2\right] = H_{k-1}\Gamma(k)$$ Then by the recurrence relation $$\left[k+1\atop 2\right] = k\left[k\atop 2\right] + \left[k\atop … Continue reading | Tagged , , , , , | Leave a comment Stirling numbers of the first kind special values proof Prove that$$\left[n\atop 1\right] = \Gamma(n)$$Use the recurrence relation$$\left[n+1\atop k\right] = n\left[n\atop k\right] + \left[n\atop k-1\right]$$This implies that for $$k=1$$$$\left[n+1\atop 1\right] = n\left[n\atop 1\right] + \left[n\atop0\right] Now use that \( \left[n\atop 0\right] = 0 … Continue reading

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