# Euler sums

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##### 1: 25.16 Mathematical Applications
###### §25.16(ii) EulerSums
Euler sums have the form … For integer $s$ ($\geq 2$), $H\left(s\right)$ can be evaluated in terms of the zeta function: … $H\left(s\right)$ is the special case $H\left(s,1\right)$ of the function …when both $H\left(s,z\right)$ and $H\left(z,s\right)$ are finite. …
##### 2: 24.14 Sums
###### §24.14 Sums
24.14.3 $\sum_{k=0}^{n}{n\choose k}E_{k}\left(h\right)E_{n-k}\left(x\right)=2(E_{n+1}% \left(x+h\right)-(x+h-1)E_{n}\left(x+h\right)),$
24.14.5 $\sum_{k=0}^{n}{n\choose k}E_{k}\left(h\right)B_{n-k}\left(x\right)=2^{n}B_{n}% \left(\tfrac{1}{2}(x+h)\right),$
For other sums involving Bernoulli and Euler numbers and polynomials see Hansen (1975, pp. 331–347) and Prudnikov et al. (1990, pp. 383–386).
##### 5: 5.7 Series Expansions
5.7.1 $\frac{1}{\Gamma\left(z\right)}=\sum_{k=1}^{\infty}c_{k}z^{k},$
5.7.5 $\psi\left(1+z\right)=\frac{1}{2z}-\frac{\pi}{2}\cot\left(\pi z\right)+\frac{1}% {z^{2}-1}+1-\gamma-\sum_{k=1}^{\infty}(\zeta\left(2k+1\right)-1)z^{2k},$ $|z|<2$, $z\neq 0,\pm 1$.
5.7.6 $\psi\left(z\right)=-\gamma-\frac{1}{z}+\sum_{k=1}^{\infty}\frac{z}{k(k+z)}=-% \gamma+\sum_{k=0}^{\infty}\left(\frac{1}{k+1}-\frac{1}{k+z}\right),$
##### 6: Bibliography F
• P. Flajolet and B. Salvy (1998) Euler sums and contour integral representations. Experiment. Math. 7 (1), pp. 15–35.
• ##### 7: 8.7 Series Expansions
8.7.1 $\gamma^{*}\left(a,z\right)=e^{-z}\sum_{k=0}^{\infty}\frac{z^{k}}{\Gamma\left(a% +k+1\right)}=\frac{1}{\Gamma\left(a\right)}\sum_{k=0}^{\infty}\frac{(-z)^{k}}{% k!(a+k)}.$
8.7.2 $\gamma\left(a,x+y\right)-\gamma\left(a,x\right)=\Gamma\left(a,x\right)-\Gamma% \left(a,x+y\right)=e^{-x}x^{a-1}\sum_{n=0}^{\infty}\frac{{\left(1-a\right)_{n}% }}{(-x)^{n}}(1-e^{-y}e_{n}(y)),$ $|y|<|x|$.
8.7.3 $\Gamma\left(a,z\right)=\Gamma\left(a\right)-\sum_{k=0}^{\infty}\frac{(-1)^{k}z% ^{a+k}}{k!(a+k)}=\Gamma\left(a\right)\left(1-z^{a}e^{-z}\sum_{k=0}^{\infty}% \frac{z^{k}}{\Gamma\left(a+k+1\right)}\right),$ $a\neq 0,-1,-2,\dots$.
8.7.4 $\gamma\left(a,x\right)=\Gamma\left(a\right)x^{\frac{1}{2}a}e^{-x}\sum_{n=0}^{% \infty}e_{n}(-1)x^{\frac{1}{2}n}I_{n+a}\left(\textstyle 2x^{1/2}\right),$ $a\neq 0,-1,-2,\dots$.
8.7.6 $\Gamma\left(a,x\right)=x^{a}e^{-x}\sum_{n=0}^{\infty}\frac{L^{(a)}_{n}\left(x% \right)}{n+1},$ $x>0$.
##### 8: 27.5 Inversion Formulas
27.5.4 $n=\sum_{d\mathbin{|}n}\phi\left(d\right)\Longleftrightarrow\phi\left(n\right)=% \sum_{d\mathbin{|}n}d\mu\left(\frac{n}{d}\right),$
##### 9: Bibliography B
• A. Basu and T. M. Apostol (2000) A new method for investigating Euler sums. Ramanujan J. 4 (4), pp. 397–419.
• ##### 10: 6.15 Sums
6.15.1 $\sum_{n=1}^{\infty}\mathrm{Ci}\left(\pi n\right)=\tfrac{1}{2}(\ln 2-\gamma),$
6.15.3 $\sum_{n=1}^{\infty}(-1)^{n}\mathrm{Ci}\left(2\pi n\right)=1-\ln 2-\tfrac{1}{2}\gamma,$