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I would like to know if the following Taylor series have a nice closed form:

$$f_k(x) = \sum_{n\geq 0}\frac{x^n}{(n+k)n!}\qquad k=1,2,3,\ldots $$ $f_1(x)=\frac{e^x-1}{x}$, but what about $f_2,f_3,f_4,\ldots$?

3 Answers 3

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Hint:

From

$$x^ke^x=\sum_{n=0}^\infty\frac{x^{k+n}}{n!}$$ one deduces

$$\int_0^xx^ke^xdx=x^{k+1}\sum_{n=0}^\infty\frac{x^n}{(n+k)n!}.$$

The antiderivative will be of the form $P(x)e^x$ where $P$ is a polynomial of degree $k$, such that $P(x)+P'(x)=x^k$.

For instance, with $k=5$,

$$\sum_{n=0}^\infty\frac{x^n}{(n+k)n!}=\frac{(x^5-5x^4+20x^3-60x^2+120x-120)e^x+120}{x^6}.$$

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    Thank you very much! You just saved my life!2017-01-23
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Claim: for any $k\in\mathbb{N}^+$, $$ \sum_{n\geq 0}\frac{x^{n}}{(n+k)n!} = \frac{p_k(x)e^x+(-1)^{k+1}k!}{x^{k+1}}\tag{1}$$ where $p_k(x)$ is a polynomial with degree $k$.

The claim is trivial for $k=1$: $\sum_{n\geq 0}\frac{x^n}{(n+1)!}=\frac{e^x-1}{x}$. In the general case, we may notice that $$ \frac{d}{dx}\left(x^{k+1}\sum_{n\geq 0}\frac{x^{n}}{(n+k)n!}\right) = \sum_{n\geq 0}\frac{x^{n+k}}{n!}=x^k e^x\tag{2}$$ so by integrating both sides we get: $$ p_k(x) = x^k-kx^{k-1}+k(k-1) x^{k-2}-k(k-1)(k-2) x^{k-3}+\ldots \tag{3}$$ or: $$ p_k(x) = k!\sum_{j=0}^{k} x^{k-j}\frac{(-1)^j}{(k-j)!}.\tag{4} $$

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\mrm{f}_{k}\pars{x} \equiv \sum_{n \geq 0}{x^{n} \over \pars{n + k}n!}:\ {\large ?} \qquad k = 1,2,3,\ldots}$

\begin{align} \mrm{f}_{k}\pars{x} & \equiv \sum_{n \geq 0}{x^{n} \over \pars{n + k}n!} = \sum_{n = 0}^{\infty}{x^{n} \over n!}\int_{0}^{1}y^{n + k - 1}\,\,\,\dd y = \int_{0}^{1}y^{k - 1}\sum_{n = 0}^{\infty}{\pars{xy}^{n} \over n!}\,\dd y = \int_{0}^{1}y^{k - 1}\expo{xy}\,\dd y \\[5mm] & \stackrel{xy\ \mapsto\ y}{=}\,\,\, {1 \over x^{k}}\int_{0}^{x}y^{k - 1}\expo{y}\,\dd y = \bbx{\ds{\left\{\begin{array}{lcrcl} \ds{\Gamma\pars{k} - \Gamma\pars{k,-x} \over \pars{-x}^{k}} & \mbox{if} & \ds{x} & \ds{\not=} & \ds{0} \\[2mm] \ds{1 \over k} & \mbox{if} & \ds{x} & \ds{=} & \ds{0} \end{array}\right.}} \end{align}

$\ds{\Gamma\pars{k}}$ and $\ds{\Gamma\pars{k,-x}}$ are the $Gamma\ Function$ and the $Incomplete\ Gamma\ Function$, respectively.