# Tag Archives: first

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 $$\sum_{n=3}^\infty ( H^{(2)}_{n-1}- (H_{n-1})^2) … Continue reading Posted in Euler sum, Striling numbers of first kind | Tagged , , , , , , , | Leave a comment ## Stirling numbers of first kind generating function Prove the following$$\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$$This implies … Continue reading | Tagged , , , , , , | Leave a comment ## Signed Stirling numbers of first kind as coefficients Signed Stirling numbers of the first kind We define the following$$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$$Which can be … Continue reading | Tagged , , , , , , | Leave a comment ## Generating number of the Stirling numbers of the first kind Prove that$$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 ## 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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