SE上有关无穷求和(欧拉和)
\begin{equation}\label{eq:n2}
\sum_{n=1}^\infty\frac{1}{n^2}
\end{equation}
的讨论。参考Different methods to compute $\sum_{k=1}^\infty\frac{1}{k^2}$ (Basel problem).
我的问题是, 如何用他们的办法求
\begin{equation}\label{eq:n3}
\sum_{n=1}^\infty\frac{1}{(n+a)(n+b)(n+c)}
\end{equation}
注意到
$$
\sum_{n=1}^\infty\frac{1}{n^3}
$$
是$\zeta(3)$并不能准确算出来(不能用已知常数表示)。
MMA模拟可知, 对不完全相同的$a,b,c$, \eqref{eq:n3}的值是可以具体计算的。
<code>max = 3; Table[{a, b, c}, {a, 0, max}, {b, 0, max}, {c, b + 1, max}]; list = Flatten[%, 2]; DeleteDuplicates[%, (Sort[#1] == Sort[#2]) &] f[a_, b_, c_] := {a, b, c, Sum[1/((n + a) (n + b) (n + c)), {n, 1, \[Infinity]}]} Apply[f, %%, {1}] // TableForm </code>
其结果如下
\begin{array}{cccl}
a & b & c & \sum\limits_{n=1}^\infty\frac{1}{(n+a)(n+b)(n+c)}\\
\hline
0 & 0 & 1 & -1+\frac{\pi ^2}{6} \\
0 & 0 & 2 & \frac{1}{24} \left(-9+2 \pi ^2\right) \\
0 & 0 & 3 & \frac{1}{54} \left(-11+3 \pi ^2\right) \\
0 & 1 & 2 & \frac{1}{4} \\
0 & 1 & 3 & \frac{7}{36} \\
0 & 2 & 3 & \frac{5}{36} \\
1 & 0 & 1 & 2-\frac{\pi ^2}{6} \\
1 & 1 & 2 & \frac{1}{6} \left(-9+\pi ^2\right) \\
1 & 1 & 3 & \frac{1}{24} \left(-17+2 \pi ^2\right) \\
1 & 2 & 3 & \frac{1}{12} \\
2 & 0 & 2 & \frac{1}{12} \left(12-\pi ^2\right) \\
2 & 1 & 2 & \frac{1}{12} \left(21-2 \pi ^2\right) \\
2 & 2 & 3 & \frac{1}{12} \left(-19+2 \pi ^2\right) \\
3 & 0 & 3 & \frac{1}{324} \left(213-18 \pi ^2\right) \\
3 & 1 & 3 & \frac{1}{36} \left(32-3 \pi ^2\right) \\
3 & 2 & 3 & \frac{1}{36} \left(61-6 \pi ^2\right) \\
\end{array}
