Σ[k=1,∞]k^n/k!の級数

次の式を \(S_n\) と置きます。$$S_n=\displaystyle\sum_{k=1}^{\infty} \frac{k^n}{k!}  \left(=\displaystyle\sum_{k=0}^{\infty} \frac{k^n}{k!}\right)$$このとき、それぞれの \(n\) における値は以下となります。
\begin{alignat}{2}
&S_1=e,  S_2=2e,  S_3=5e,  S_4=15e\\
&\\
&S_5=52e,  S_6=203e,  S_7=877e,  S_8=4140e\\
\end{alignat}
また、上記の計算の要領で次式を得ます。$$(α)  S_n=\displaystyle\sum_{k=0}^{n-1} {}_{n-1}\mathrm{C}_k S_k$$


別の表し方として \(\displaystyle f_1(x)=e^x=\displaystyle\sum_{k=0}^{\infty} \frac{x^k}{k!},\,\,f_n(x)=\{xf_{n-1}(x)\}’\) と置くと$$(β)  S_n=\displaystyle\sum_{k=1}^{\infty} \frac{k^n}{k!}=f_n(1)$$







<証明>

\((α)\) \((1)\) \(n=1\) のとき$$S_1=\displaystyle\sum_{k=1}^{\infty}\frac{k}{k!}=\displaystyle\sum_{k=1}^{\infty}\frac{1}{(k-1)!}=\displaystyle\sum_{k=0}^{\infty} \frac{1}{k!}=e$$



\((2)\) \(n=2\) のとき
\begin{alignat}{2}
S_2&=\displaystyle\sum_{k=1}^{\infty}\frac{k^2}{k!}=\displaystyle\sum_{k=1}^{\infty}\frac{k}{(k-1)!}=\displaystyle\sum_{k=0}^{\infty} \frac{k+1}{k!}\\
&=\displaystyle\sum_{k=0}^{\infty} \frac{k}{k!}+ \displaystyle\sum_{k=0}^{\infty} \frac{1}{k!}=S_1+e=e+e=2e\\
\end{alignat}




\((3)\) \(n=3\) のとき
\begin{alignat}{2}
S_3&=\displaystyle\sum_{k=1}^{\infty}\frac{k^3}{k!}=\displaystyle\sum_{k=1}^{\infty}\frac{k^2}{(k-1)!}=\displaystyle\sum_{k=0}^{\infty} \frac{(k+1)^2}{k!}\\
&=\displaystyle\sum_{k=0}^{\infty} \frac{k^2+2k+1}{k!}=\displaystyle\sum_{k=0}^{\infty} \frac{k^2}{k!}+2\displaystyle\sum_{k=0}^{\infty} \frac{k}{k!}+\displaystyle\sum_{k=0}^{\infty} \frac{1}{k!}\\
&=S_2+2S_1+e=2e+2e+e=5e
\end{alignat}




\((4)\) \(n=4\) のとき
\begin{alignat}{2}
S_4&=\displaystyle\sum_{k=1}^{\infty}\frac{k^4}{k!}=\displaystyle\sum_{k=1}^{\infty}\frac{k^3}{(k-1)!}=\displaystyle\sum_{k=0}^{\infty} \frac{(k+1)^3}{k!}\\
&=\displaystyle\sum_{k=0}^{\infty} \frac{k^3+3k^2+3k+1}{k!}=\displaystyle\sum_{k=0}^{\infty} \frac{k^3}{k!}+3\displaystyle\sum_{k=0}^{\infty} \frac{k^2}{k!}+3\displaystyle\sum_{k=0}^{\infty} \frac{k}{k!}+\displaystyle\sum_{k=0}^{\infty} \frac{1}{k!}\\
&=S_3+3S_2+3S_1+e=5e+6e+3e+e=15e
\end{alignat}





\((5)\) \(n=5\) のとき
\begin{alignat}{2}
S_5&=\displaystyle\sum_{k=1}^{\infty}\frac{k^5}{k!}=\displaystyle\sum_{k=1}^{\infty}\frac{k^4}{(k-1)!}=\displaystyle\sum_{k=0}^{\infty} \frac{(k+1)^4}{k!}\\
&=\displaystyle\sum_{k=0}^{\infty} \frac{k^4+4k^3+6k^2+4k+1}{k!}\\
&=\displaystyle\sum_{k=0}^{\infty} \frac{k^4}{k!}+4\displaystyle\sum_{k=0}^{\infty} \frac{k^3}{k!}+6\displaystyle\sum_{k=0}^{\infty} \frac{k^2}{k!}+4\displaystyle\sum_{k=0}^{\infty} \frac{k}{k!}+\displaystyle\sum_{k=0}^{\infty} \frac{1}{k!}\\
&=S_4+4S_3+6S_2+4S_1+e=15e+20e+12e+4e+e=52e
\end{alignat}




\((6)\) \(n=6\) のとき
\begin{alignat}{2}
S_6&=\displaystyle\sum_{k=1}^{\infty}\frac{k^6}{k!}=\displaystyle\sum_{k=1}^{\infty}\frac{k^5}{(k-1)!}=\displaystyle\sum_{k=0}^{\infty} \frac{(k+1)^5}{k!}\\
&=\displaystyle\sum_{k=0}^{\infty} \frac{k^5+5k^4+10k^3+10k^2+5k+1}{k!}\\
&=\displaystyle\sum_{k=0}^{\infty} \frac{k^5}{k!}+5\displaystyle\sum_{k=0}^{\infty} \frac{k^4}{k!}+10\displaystyle\sum_{k=0}^{\infty} \frac{k^3}{k!}+10\displaystyle\sum_{k=0}^{\infty} \frac{k^2}{k!}+5\displaystyle\sum_{k=0}^{\infty} \frac{k}{k!}+\displaystyle\sum_{k=0}^{\infty} \frac{1}{k!}\\
&=S_5+5S_4+10S_3+10S_2+5S_1+e=52e+75e+50e+20e+5e+e=203e
\end{alignat}





\((7)\) \(n=7\) のとき
\begin{alignat}{2}
S_7&=\displaystyle\sum_{k=1}^{\infty}\frac{k^7}{k!}=\displaystyle\sum_{k=1}^{\infty}\frac{k^6}{(k-1)!}=\displaystyle\sum_{k=0}^{\infty} \frac{(k+1)^6}{k!}\\
&=\displaystyle\sum_{k=0}^{\infty} \frac{k^6+6k^5+15k^4+20k^3+15k^2+6k+1}{k!}\\
&=\displaystyle\sum_{k=0}^{\infty} \frac{k^6}{k!}+6\displaystyle\sum_{k=0}^{\infty} \frac{k^5}{k!}+15\displaystyle\sum_{k=0}^{\infty} \frac{k^4}{k!}+20\displaystyle\sum_{k=0}^{\infty} \frac{k^3}{k!}+15\displaystyle\sum_{k=0}^{\infty} \frac{k^2}{k!}+6\displaystyle\sum_{k=0}^{\infty} \frac{k}{k!}+\displaystyle\sum_{k=0}^{\infty} \frac{1}{k!}\\
&=S_6+6S_5+15S_4+20S_3+15S_2+6S_1+e\\
&=203e+312e+225e+100e+30e+6e+e=877e
\end{alignat}





\((8)\) \(n=8\) のとき
\begin{alignat}{2}
S_8&=\displaystyle\sum_{k=1}^{\infty}\frac{k^8}{k!}=\displaystyle\sum_{k=1}^{\infty}\frac{k^7}{(k-1)!}=\displaystyle\sum_{k=0}^{\infty} \frac{(k+1)^7}{k!}\\
&=\displaystyle\sum_{k=0}^{\infty} \frac{k^7+7k^6+21k^5+35k^4+35k^3+21k^2+7k+1}{k!}\\
&=\displaystyle\sum_{k=0}^{\infty} \frac{k^7}{k!}+7\displaystyle\sum_{k=0}^{\infty} \frac{k^6}{k!}+21\displaystyle\sum_{k=0}^{\infty} \frac{k^5}{k!}+35\displaystyle\sum_{k=0}^{\infty} \frac{k^4}{k!}+35\displaystyle\sum_{k=0}^{\infty} \frac{k^3}{k!}+21\displaystyle\sum_{k=0}^{\infty} \frac{k^2}{k!}+7\displaystyle\sum_{k=0}^{\infty} \frac{k}{k!}+\displaystyle\sum_{k=0}^{\infty} \frac{1}{k!}\\
&=S_7+7S_6+21S_5+35S_4+35S_3+21S_2+7S_1+e\\
&=877e+1421e+1092e+525e+175e+42e+7e+e=4140e
\end{alignat}





\begin{alignat}{2}
(9)  S_n&=\displaystyle\sum_{k=1}^{\infty}\frac{k^n}{k!}=\displaystyle\sum_{k=1}^{\infty}\frac{k^{n-1}}{(k-1)!}=\displaystyle\sum_{k=0}^{\infty} \frac{(k+1)^{n-1}}{k!}\\
&=\displaystyle\sum_{k=0}^{\infty} \frac{1}{k!}\left(\displaystyle\sum_{l=0}^{n-1} {}_{n-1}\mathrm{C}_l k^l\right)\\
&=\displaystyle\sum_{k=0}^{\infty} \frac{1}{k!}({}_{n-1}\mathrm{C}_{n-1} k^{n-1}+{}_{n-1}\mathrm{C}_{n-2} k^{n-2}+ \cdots +{}_{n-1}\mathrm{C}_2 k^2+{}_{n-1}\mathrm{C}_1 k+{}_{n-1}\mathrm{C}_0 )\\
&={}_{n-1}\mathrm{C}_{n-1}\displaystyle\sum_{k=0}^{\infty} \frac{k^{n-1}}{k!}+{}_{n-1}\mathrm{C}_{n-2}\displaystyle\sum_{k=0}^{\infty} \frac{k^{n-2}}{k!}+ \cdots +{}_{n-1}\mathrm{C}_{2}\displaystyle\sum_{k=0}^{\infty} \frac{k^{2}}{k!}+{}_{n-1}\mathrm{C}_{1}\displaystyle\sum_{k=0}^{\infty} \frac{k}{k!}+{}_{n-1}\mathrm{C}_0\displaystyle\sum_{k=0}^{\infty} \frac{1}{k!}\\
&={}_{n-1}\mathrm{C}_{n-1}S_{n-1}+{}_{n-1}\mathrm{C}_{n-2}S_{n-2}+ \cdots +{}_{n-1}\mathrm{C}_{2}S_2+{}_{n-1}\mathrm{C}_{1}S_1+{}_{n-1}\mathrm{C}_0 S_0\\
&=\displaystyle\sum_{k=0}^{n-1} {}_{n-1}\mathrm{C}_k S_k
\end{alignat}以上より$$S_n=\displaystyle\sum_{k=0}^{n-1} {}_{n-1}\mathrm{C}_k S_k$$









\((β)\) 予め、いくつかの微分計算を行います。

\(e^x\) に \(x\) を掛けて微分、\(x\) を掛けて微分…を繰り返します。$$(A)  (xe^x)’=e^x+xe^x=(x+1)e^x$$
\begin{alignat}{2}
(B)  \{x(x+1)e^x\}’&=\{(x^2+x)e^x\}’=(2x+1)e^x+(x^2+x)e^x\\
&=(x^2+3x+1)e^x\\
\end{alignat}
\begin{alignat}{2}
(C)  \{x(x^2+3x+1)e^x\}’&=\{(x^3+3x^2+x)e^x\}’\\
&=(3x^2+6x+1)e^x+(x^3+3x^2+x)e^x\\
&=(x^3+6x^2+7x+1)e^x\\
\end{alignat}
\begin{alignat}{2}
(D)  \{x(x^3+6x^2+7x+1)e^x\}’&=\{(x^4+6x^3+7x^2+x)e^x\}’\\
&=(4x^3+18x^2+14x+1)e^x+(x^4+6x^3+7x^2+x)e^x\\
&=(x^4+10x^3+25x^2+15x+1)e^x\\
\end{alignat}
\begin{alignat}{2}
(E)  \{x(x^4+10x^3+25x^2+15x+1)e^x\}’&=\{(x^5+10x^4+25x^3+15x^2+x)e^x\}’\\
&=(5x^4+40x^3+75x^2+30x+1)e^x+(x^5+10x^4+25x^3+15x^2+x)e^x\\
&=(x^5+15x^4+65x^3+90x^2+31x+1)e^x\\
\end{alignat}
\begin{alignat}{2}
(F)  \{x(x^5+15x^4+65x^3+90x^2+31x+1)e^x\}’&=\{(x^6+15x^5+65x^4+90x^3+31x^2+x)e^x\}’\\
&=(6x^5+75x^4+260x^3+270x^2+62x+1)e^x+(x^6+15x^5+65x^4+90x^3+31x^2+x)e^x\\
&=(x^6+21x^5+140x^4+350x^3+301x^2+63x+1)e^x\\
\end{alignat}
\begin{alignat}{2}
(G)  \{x(x^6+21x^5+140x^4+350x^3+301x^2+63x+1)e^x\}’&=\{(x^7+21x^6+140x^5+350x^4+301x^3+63x^2+x)e^x\}’\\
&=(7x^6+126x^5+700x^4+1400x^3+903x^2+126x+1)e^x+(x^7+21x^6+140x^5+350x^4+301x^3+63x^2+x)e^x\\
&=(x^7+28x^6+266x^5+1050x^4+1701x^3+966x^2+127x+1)e^x
\end{alignat}

上記 \((A)\) から\((G)\) を用いて \(S_1\) から \(S_8\) までを計算します。

まず \(f_1(x)\) から \(f_8(x)\) については

\begin{alignat}{2}
&f_2(x)=\{xf_1(x)\}’=(xe^x)’=(x+1)e^x\\
&\\
&f_3(x)=\{xf_2(x)\}’=\{x(x+1)e^x\}’=(x^2+3x+1)e^x\\
&\\
&f_4(x)=\{xf_3(x)\}’=\{x(x^2+3x+1)e^x\}’=(x^3+6x^2+7x+1)e^x\\
&\\
&f_5(x)=\{xf_4(x)\}’=\{x(x^3+6x^2+7x+1)e^x\}’=(x^4+10x^3+25x^2+15x+1)e^x\\
&\\
&f_6(x)=\{xf_5(x)\}’=\{x(x^4+10x^3+25x^2+15x+1)e^x\}’=(x^5+15x^4+65x^3+90x^2+31x+1)e^x\\
&\\
&f_7(x)=\{xf_6(x)\}’=\{x(x^5+15x^4+65x^3+90x^2+31x+1)e^x\}’=(x^6+21x^5+140x^4+350x^3+301x^2+63x+1)e^x\\
&\\
&f_8(x)=\{xf_7(x)\}’=\{x(x^6+21x^5+140x^4+350x^3+301x^2+63x+1)e^x\}’=(x^7+28x^6+266x^5+1050x^4+1701x^3+966x^2+127x+1)e^x\\
\end{alignat}
であるので、全て \(x=1\) とすれば
\begin{alignat}{2}
&S_1=f_1(1)=e,  S_2=f_2(1)=2e\\
&\\
&S_3=f_3(1)=5e,  S_4=f_4(1)=15e\\
&\\
&S_5=f_5(1)=52e,  S_6=f_6(1)=203e\\
&\\
&S_7=f_7(1)=877e,  S_8=f_8(1)=4140e\\
\end{alignat}




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