Category Archives: Striling numbers of first kind

General formula for an integral involving powers of logarithms

$$\int\limits_0^1 \dfrac{\log^{m} (1+x)\log^n x}{x}\; dx = (-1)^{n+1}(n!) (m)! \sum_{\{m-1\}} \sum_{k=1}^\infty \frac{(-1)^k}{k^{n+2}} \prod^{l’}_{j=1}\frac{(-1)^{i_j}}{(i_j)!} \left( \frac{H_{k-1}^{(r_j)}}{r_j}\right)^{i_j}$$ $$\textit{solution}$$ 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

Posted in Euler sum, Harmonic numbers, stirling, Striling numbers of first kind | Tagged , , , , , | 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

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

Posted in Signed Stirling numbers of first kind, Striling numbers of first kind | 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

Posted in Signed Stirling numbers of first kind, Striling numbers of first kind | 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 | Tagged , , , , , , , , | 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

Posted in Harmonic numbers, Striling numbers of first kind | 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

Posted in Striling numbers of first kind | Tagged , , , , , , | Leave a comment