一个与多项式分拆有关的级数题

求证：对于$\forall k\in N_+$，必有\[\sum\limits_{n = 1}^\infty  {\frac{{{n^k}}}{{n!}}}\]是$e$的整数倍.

证明.先证明一个引理：对$\forall k\in N_+$，均有${n^k} = {a_0}n + {a_1}n\left( {n - 1} \right) + {a_2}n\left( {n - 1} \right)\left( {n - 2} \right) +  \cdots  + {a_k}n\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( {n - k+1} \right)$成立，其中$a_1+a_2+\ldots+a_i+\ldots+a_{k-1}(i=0,1,2,\ldots,k)$为整数（事实上，$a_i$均为正整数，$i\in N^+$）.

证明.写得规范点，我们有：\[{x^k} = {a_0}x + {a_1}x\left( {x - 1} \right) + {a_2}x\left( {x - 1} \right)\left( {x - 2} \right) +  \cdots  + {a_{k-1}}x\left( {x - 1} \right)\left( {x - 2} \right) \cdots \left( {x - k+1} \right).\]

比较两边$x^k$的系数得知$a_{k-1}=1$，即

\begin{align*}&x\left( {x - 1} \right)\left( {x - 2} \right) \cdots \left( {x - k + 1} \right) = {x^k} - {a_{k - 2}}x\left( {x - 1} \right)\left( {x - 2} \right) \cdots \left( {x - k + 2} \right) \\&-  \cdots  - {a_2}x\left( {x - 1} \right)\left( {x - 2} \right) - {a_1}x\left( {x - 1} \right) - {a_0}x.\end{align*}

 

再比较两边$x^{k-1}$的系数得$a_{k-2}=\frac{k(k-1)}{2}$，由此

\[\frac{{{{\left( {1 + 2 +  \cdots  + \left( {k - 1} \right)} \right)}^2} - \left( {{1^2} + {2^2} +  \cdots  + {{\left( {k - 1} \right)}^2}} \right)}}{2} = \left( {1 + 2 +  \cdots  + \left( {k - 2} \right)} \right){a_{k - 2}} - {a_{k - 3}}.\]故\[{a_{k - 3}} = \frac{{\left( {3k - 5} \right)\left( {k - 2} \right)\left( {k - 1} \right)k}}{{24}}.\]

一般地，考察等式两边$x^n$的系数我们有

\begin{align*}&{\left( { - 1} \right)^{k - n}}\sum\limits_{1 \le {k_1} < {k_2} <  \cdots  < {k_{k - n}} \le k - 1} {{k_1}{k_2} \cdots {k_{k - n}}}  = {\left( { - 1} \right)^{k - n}}{a_{k - 2}}\sum\limits_{1 \le {k_1} < {k_2} <  \cdots  < {k_{k - n - 1}} \le k - 2} {{k_1}{k_2} \cdots {k_{k - n - 1}}} \\&+ {\left( { - 1} \right)^{k - n - 1}}{a_{k - 3}}\sum\limits_{1 \le {k_1} < {k_2} <  \cdots  < {k_{k - n - 2}} \le k - 3} {{k_1}{k_2} \cdots {k_{k - n - 2}}}  +  \cdots  + {\left( { - 1} \right)^2}{a_n}\sum\limits_{1 \le {k_1} \le n} {{k_1}} {x^n} + {\left( { - 1} \right)^1}{a_{n - 1}}.\end{align*}

依此递推式得$a_0=1.$且$a_i,i=0,1,\ldots,k-2$均为正整数.

回到原题，我们有

\begin{align*}\sum\limits_{n = 1}^\infty  {\frac{{{n^k}}}{{n!}}}  &= \sum\limits_{n = 1}^\infty  {\frac{{{a_0}n + {a_1}n\left( {n - 1} \right) + {a_2}n\left( {n - 1} \right)\left( {n - 2} \right) +  \cdots  + {a_{k - 1}}n\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( {n - k + 1} \right)}}{{n!}}} \\&= {a_0}\sum\limits_{n = 1}^\infty  {\frac{1}{{\left( {n - 1} \right)!}}}  + {a_1}\sum\limits_{n = 2}^\infty  {\frac{1}{{\left( {n - 2} \right)!}}}  + {a_2}\sum\limits_{n = 3}^\infty  {\frac{1}{{\left( {n - 3} \right)!}}}  +  \cdots  + {a_{k - 1}}\sum\limits_{n = k}^\infty  {\frac{1}{{\left( {n - k} \right)!}}} \\&= \left( {{a_0} + {a_1} + {a_2} +  \cdots {a_{k - 1}}} \right)e = Ne.\end{align*}