Exponentials for us

We know that $\displaystyle \sum_{k=0}^\infty \frac {k^n}{k!} = B_n e$ , where $B_n$ is the $n$ th Bell number for $n$ is a non-negative integer. And Bell number can be generated by $\displaystyle B_n = \sum_{k=0}^{n-1}B_k \binom {n-1}k$ . We can easily get from $B_0 = \displaystyle \frac 1e \sum_{k=0}^\infty \frac 1{k!} = \frac ee = 1$ . Then we have:

$\begin{aligned} B_1 & = B_0 \dbinom {1-1}0 = 1 \\ B_2 & = B_0 \dbinom {2-1}0 + B_1 \dbinom {2-1}1 = 1+1 = 2 \\ B_3 & = B_0 \dbinom {3-1}0 + B_1 \dbinom {3-1}1 + B_2 \dbinom {3-1}2 = 1 + 2 + 2 = 5 \\ B_4 & = B_0 \dbinom {4-1}0 + B_1 \dbinom {4-1}1 + B_2 \dbinom {4-1}2 + B_3 \dbinom {4-1}3 = 1 + 3 + 6 + 5 = 15 \end{aligned}$

Therefore, we have $\dfrac {\sum_{k=1}^\infty \frac {k^4}{k!}}{\sum_{k=1}^\infty \frac {k^3}{k!}} = \dfrac {\sum_{k=0}^\infty \frac {k^4}{k!}}{\sum_{k=0}^\infty \frac {k^3}{k!}} = \dfrac {B_4e}{B_3n} = \dfrac {15}5 = \boxed{3}$ .

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Alternative solution

$\begin{aligned} \sum_{k=0}^\infty \frac 1{k!} & = e \\ \sum_{k=1}^\infty \frac k{k!} & = \sum_{k=0}^\infty \frac k{k!} = \sum_{k=0}^\infty \frac {k+1}{(k+1)!} = \sum_{k=0}^\infty \frac 1{k!} = e \\ \sum_{k=1}^\infty \frac {k^2}{k!} & = \sum_{k=0}^\infty \frac {k^2}{k!} = \sum_{k=0}^\infty \frac {(k+1)^2}{(k+1)!} = \sum_{k=0}^\infty \frac {k+1}{k!} = 2e \\ \sum_{k=1}^\infty \frac {k^3}{k!} & = \sum_{k=0}^\infty \frac {k^3}{k!} = \sum_{k=0}^\infty \frac {(k+1)^3}{(k+1)!} = \sum_{k=0}^\infty \frac {(k+1)^2}{k!} = \sum_{k=0}^\infty \frac {k^2+2k+1}{k!} = 5e \\ \sum_{k=1}^\infty \frac {k^4}{k!} & = \sum_{k=0}^\infty \frac {k^4}{k!} = \sum_{k=0}^\infty \frac {(k+1)^4}{(k+1)!} = \sum_{k=0}^\infty \frac {(k+1)^3}{k!} = \sum_{k=0}^\infty \frac {k^3+3k^2+3k+1}{k!} = 15e \end{aligned}$

Therefore, we have $\dfrac {\sum_{k=1}^\infty \frac {k^4}{k!}}{\sum_{k=1}^\infty \frac {k^3}{k!}} = \dfrac {15e}{5e} = \boxed{3}$ .

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Alternative solution

$\begin{aligned} \sum_{k=0}^\infty \frac {x^k}{k!} & = e^x & \small \color{#3D99F6} \implies x=1 \implies \sum_{k=0}^\infty \frac 1{k!} = e \\ \frac d{dx}\sum_{k=0}^\infty \frac {x^k}{k!} & = \frac d{dx} e^x \\ \sum_{k=0}^\infty \frac {kx^{k-1}}{k!} & = e^x & \small \color{#3D99F6} \implies x=1 \implies \sum_{k=0}^\infty \frac k{k!} = e \\ \sum_{k=0}^\infty \frac {kx^k}{k!} & = xe^x \\ \frac d{dx} \sum_{k=0}^\infty \frac {kx^k}{k!} & = \frac d{dx} xe^x \\ \sum_{k=0}^\infty \frac {k^2x^{k-1}}{k!} & = e^x + xe^x & \small \color{#3D99F6} \implies x=1 \implies \sum_{k=0}^\infty \frac {k^2}{k!} = 2e \\ \sum_{k=0}^\infty \frac {k^2x^k}{k!} & = xe^x + x^2e^x \\ \frac d{dx} \sum_{k=0}^\infty \frac {k^2x^k}{k!} & = \frac d{dx} \left(xe^x + x^2e^x\right) \\ \sum_{k=0}^\infty \frac {k^3x^{k-1}}{k!} & = e^x + 3xe^x + x^2e^x & \small \color{#3D99F6} \implies x=1 \implies \sum_{k=0}^\infty \frac {k^3}{k!} = 5e \\ \sum_{k=0}^\infty \frac {k^3x^k}{k!} & = xe^x + 3x^2e^x + x^3e^x \\ \frac d{dx} \sum_{k=0}^\infty \frac {k^3x^k}{k!} & = \frac d{dx} \left(xe^x + 3x^2e^x + x^3e^x\right) \\ \sum_{k=0}^\infty \frac {k^4x^{k-1}}{k!} & = e^x + 7xe^x + 6x^2e^x + x^3e^x & \small \color{#3D99F6} \implies x=1 \implies \sum_{k=0}^\infty \frac {k^4}{k!} = 15e \end{aligned}$

Therefore, we have $\dfrac {\sum_{k=1}^\infty \frac {k^4}{k!}}{\sum_{k=1}^\infty \frac {k^3}{k!}} = \dfrac {15e}{5e} = \boxed{3}$ .

$\begin{aligned} & S_1 =\displaystyle\sum_{n=1}^{\infty} \frac{n^4}{n!} = \frac{n^3}{(n-1)!} =\frac{(n^3-1)}{(n-1)!} + \frac{1}{(n-1)!} \\& \quad =\displaystyle\sum_{n=1}^{\infty} \frac{n^2}{(n-2)! } + \frac{n}{(n-2)!} + \frac{2}{(n-1)!} \\&\quad =\displaystyle\sum_{n=1}^{\infty} \frac{(n+2)(n-2)}{(n-2)!} +\frac{4}{(n-1)!} + \frac{1}{(n-2)!}+ \frac{3}{(n-1)!} \\& \quad =\displaystyle\sum_{n=1}^{\infty} \frac{(n-3)}{(n-3)! } + \frac{3}{(n-1)!}+ \frac{8}{(n-1)!}\\& \quad =\displaystyle\sum_{n=1}^{\infty} \frac{4}{(n-3)!} + \frac{11}{(n-1)!} = 15e \end{aligned}$ $\begin{aligned} S_2 = & \displaystyle\sum_{n=1}^{\infty}\frac {n^3}{n!} = \displaystyle\sum_{n=1}^{\infty}\frac {n^2-1+1}{(n-1)!} =\displaystyle\sum_{n=1}^{\infty} \frac {n+1}{(n-2)!} + \frac {1}{(n-1)!} \\& =\displaystyle\sum_{n=1}^{\infty}\frac{1}{(n-3)!} + \frac {3}{(n-2)!} + \frac {1}{(n-1)!}= 5e\end{aligned}$

Thus $\frac{S_1}{S_2} = \frac{15e}{5e} = 3$ .