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Is there a way to rewrite an equal expression to the following

$$\sum^{z}_{n=0}n\frac{x^{z-n}}{(z-n)!}$$

but without the sum?

$z$ is integer and $x\geq 0$.

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The answer is no. By making the change of avriable $k=z-n$ and splitting the sum into two parts you get $(z-x)\sum_{k=1}^{z-1} \frac {x^{k}} {k!} +\frac {zx^{z}} {z!}$ from which you cannot get rid of the sum.

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  • $\begingroup$ Thank you @Kavi Rama Murthy, but you gave the answer. I just use the Gamma function for the last sum and I get the expression. $\endgroup$ – Y.L Jan 4 at 0:28
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Sticking to a more standard notation ($z$ normally denotes complex, not integral, number)

\begin{align} \sum_{k=0}^n k \frac{x^{n-k}}{(n-k)!} &= \sum_{k=0}^{n} (n-k) \frac{x^k}{k!} = n\sum_{k=0}^n \frac{x^k}{k!} - \sum_{k=0}^n x\frac{d}{dx} \frac{x^k}{k!}\\ &= nP_n(x) - x\frac{d}{dx}P_n(x) \end{align}

where $P_n(x) = \sum_{k=0}^n \frac{x^k}{k!}$ is the partial sum of the exponential's Taylor series. But as can be seen here, the incomplete gamma function $\Gamma(n,x) = \int_x^{\infty} x^{n-1}e^{-x}\;dx$ may be expressed as

$$\Gamma(n,x) = (n-1)! e^{-x} \sum_{k=0}^{n-1}\frac{x^k}{k!}$$ hence, $$P_n(x) = \frac{\Gamma(n+1,x)}{n!} e^x$$ We may then compute

\begin{align} \frac{d}{dx}P_n(x) &= P_n(x) \frac{d}{dx} \ln P_n(x) = P_n(x) \left[\,\frac{\frac{\partial}{\partial{x}}\Gamma(n+1,x)}{\Gamma(n+1,x)} + 1\,\right]\\ &= \frac{\Gamma(n+1,x)}{n!} e^x \left[\,\frac{-x^ne^{-x}}{\Gamma(n+1,x)} + 1\,\right]\\ &= \frac{\Gamma(n+1,x)}{n!} e^x - \frac{x^n}{n!} \end{align}

Solving and simplifying, we get

\begin{align} \sum_{k=0}^n k \frac{x^{n-k}}{(n-k)!} &= nP_n(x) - x\frac{d}{dx}P_n(x)\\ &= \frac{1}{n!} \left[\,(n-x)\Gamma(n+1,x)e^x + x^{n+1}\,\right] \end{align}

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