# Tag Archives: sum

$$\tag{1}1+2+3+4+\cdots =^?-\frac{1}{12}$$ There are many problems with that equality. Can a summation of positive integers lead to a negative quantity ? The equality implies that $$\sum_{k=1}^\infty k = -\frac{1}{12}$$ But we already know that $$S_N = \sum_{k=1}^Nk = \frac{N(N+1)}{2} \to … Continue reading Posted in Zeta function | | Leave a comment ## Symmetry formula for Generalized Linear Euler sums$$\sum_{k=1}^\infty \frac{H^{(p)}_k}{k^q}+\sum_{k=1}^\infty \frac{H^{(q)}_k}{k^p} =\zeta(p)\zeta(q)+\zeta(p+q)\textit{proof}$$Take the leftmost series and swap the finite and infinite sums$$\sum_{i=1}^\infty \,\sum_{k=i}^\infty\frac{1}{i^p} \frac{1}{k^q}=\sum_{i=1}^\infty \,\sum_{k=1}^\infty\frac{1}{i^p} \frac{1}{k^q}-\sum_{i=1}^\infty\frac{1}{i^p} \,\sum_{k=1}^{i-1} \frac{1}{k^q}$$The second sum can be written as$$\sum_{i=1}^\infty\frac{1}{i^p} \,\sum_{k=1}^{i-1} \frac{1}{k^q} = \sum_{i=1}^\infty\frac{1}{i^p} \,\left(\sum_{k=1}^{i} \frac{1}{k^q}-\frac{1}{i^p}\right)$$By separating and … Continue reading Posted in Euler sum | Tagged , , , , , | Leave a comment ## Integral representation of generalized Euler sums$$\sum_{k=1}^\infty\frac{H^{(p)}_k}{k^q} = \zeta(p)\zeta(q) +(-1)^{p}\frac{1}{ (p-1)!}\int^1_0\frac{\mathrm{Li}_q(x)\log(x)^{p-1}}{1-x}\,dx\textit{proof}$$Note that$$\psi_0(a+1)= \int^1_0\frac{1-x^a}{1-x}\,dx$$By differentiating with respect to $a$ , $p$ times we have$$\psi_p(a+1) = \frac{\partial}{\partial a^p}\int^1_0\frac{1-x^a}{1-x}\,dx\psi_p(a+1) = -\int^1_0\frac{x^a\log(x)^{p}}{1-x}\,dx$$Let $a =k$$$\psi_{p-1}(k+1) = -\int^1_0\frac{x^k\log(x)^{p-1}}{1-x}\,dx$$Use the relation to polygamma$$H^{(p)}_k … Continue reading

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## Nonlinear Euler sums using Nielsen formula

According to Nielsen we have the following : If $$f(x)= \sum_{n= 0}^\infty a_n x^n$$ Then we have the following $$\tag{1}\int^1_0 f(xt)\, \mathrm{Li}_2(t)\, dx=\frac{\pi^2}{6x}\int^x_0 f(t)\, dt -\frac{1}{x}\sum_{n=1}^\infty \frac{a_{n-1} H_{n}}{n^2}x^n$$ Now let $a_n = H_n$ then we have the … Continue reading

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