Closed form of $\sum_{n = 1}^{\infty} \frac{n^{n - k}}{e^{n} \cdot n!}$

When seeing this question I noticed that $$\sum_{n = 1}^{\infty} \frac{n^{n - 2}}{e^{n} \cdot n!} = \frac{1}{2}.$$ I don't know how to show this, I tried finding a power series that matches that but no avail. Hints are very much appreciated.

But this can be generalised: Define $$S_{k}(x) := \sum_{n = 1}^{\infty} \frac{n^{n - k}}{x^{n} \cdot n!}.$$ WolframAlpha shows e.g. that $$S_1(e) = 1$$ and $$S_0(x) = - \frac{W(-x^{-1})}{1 + W(-x^{-1})},$$ where $$W$$ denotes the Lambert W-function. Is there any closed form for this sum or a special $$k$$ or $$x$$ beyond those results?

Somebody attempted to answer this using the Lagrange inversion theorem. I didn't work out completely but looked quite promising.

Computing the Sums

Using the Taylor series for the Lambert W function derived in this answer, we see that $$\newcommand{\W}{\operatorname{W}} -\W(-x)=\sum_{n=1}^\infty\frac{n^{n-1}}{n!}x^n\tag1$$ Define $$u(x)=-\W(-x)$$, then we have \begin{align} 1&=u(1/e)\tag{2a}\\[3pt] x&=u(x)\,e^{-u(x)}\tag{2b}\\ \frac{\mathrm{d}x}x&=\frac{1-u(x)}{u(x)}\,\mathrm{d}u(x)\tag{2c} \end{align}

Define $$u_1=u$$ and $$u_{k+1}(x)=\int_0^xu_k(t)\,\frac{\mathrm{d}t}{t}\tag3$$ Then $$u_k(x)=\sum_{n=1}^\infty\frac{n^{n-k}}{n!}x^n\tag4$$ We will show that $$u_k(x)=P_k(u(x))\tag5$$ where $$P_k(x)$$ is a degree $$k$$ polynomial with no constant term. It is true for $$k=1$$ with $$P_1(x)=x$$. Assume it is true for $$k$$, then $$(3)$$ and $$\text{(2c)}$$ imply \begin{align} u_{k+1}(x) &=\int_0^xP_k(u(t))\,\frac{\mathrm{d}t}t\\ &=\int_0^{u(x)}P_k(u)\,\frac{1-u}u\,\mathrm{d}u\\[9pt] &=P_{k+1}(u(x))\tag6 \end{align} Thus, $$(5)$$ is true for $$k+1$$.

Equation $$(6)$$ enables us to compute $$P_{k+1}(x)=\int_0^xP_k(t)\frac{1-t}t\,\mathrm{d}t\tag7$$ The initial part of the sequence of polynomials is \begin{align} P_1(x)&=x\\ P_2(x)&=x-\tfrac12x^2\\ P_3(x)&=x-\tfrac34x^2+\tfrac16x^3\\ P_4(x)&=x-\tfrac78x^2+\tfrac{11}{36}x^3-\tfrac1{24}x^4\\ P_5(x)&=x-\tfrac{15}{16}x^2+\tfrac{85}{216}x^3-\tfrac{25}{288}x^4+\tfrac1{120}x^5\\ P_6(x)&=x-\tfrac{31}{32}x^2+\tfrac{575}{1296}x^3-\tfrac{415}{3456}x^4+\tfrac{137}{7200}x^5-\tfrac1{720}x^6\\ P_7(x)&=x-\tfrac{63}{64}x^2+\tfrac{3661}{7776}x^3-\tfrac{5845}{41472}x^4+\tfrac{12019}{432000}x^5-\tfrac{49}{14400}x^6+\tfrac1{5040}x^7\\ P_8(x)&=x{-}\tfrac{127}{128}x^2{+}\tfrac{22631}{46656}x^3{-}\tfrac{76111}{497664}x^4{+}\tfrac{874853}{25920000}x^5{-}\tfrac{13489}{2592000}x^6{+}\tfrac{121}{235200}x^7{-}\tfrac1{40320}x^8 \end{align}\tag8 Applying $$\text{(2a)}$$, $$(4)$$, and $$(5)$$, we get $$\sum_{n=1}^\infty\frac{n^{n-k}}{e^nn!}=P_k(1)\tag9$$

Table of Sums $$\begin{array}{c|c} k&\sum\limits_{n=1}^\infty\frac{n^{n-k}}{e^nn!}\\\hline 1&1\\ 2&\frac12\\ 3&\frac5{12}\\ 4&\frac7{18}\\ 5&\frac{1631}{4320}\\ 6&\frac{96547}{259200}\\ 7&\frac{40291823}{108864000}\\ 8&\frac{16870575007}{45722880000} \end{array}\tag{10}$$

Recursion for the Coefficients of $$\boldsymbol{P_k}$$

Let $$P_k(x)=\sum_{j=1}^\infty(-1)^{j-1}a_{k,j}x^j\tag{11}$$ where $$a_{k,1}=1$$ and $$a_{k,j}=0$$ for $$j\gt k$$. Applying $$(7)$$ to $$(11)$$ gives \begin{align} a_{k,j} &=\frac{a_{k-1,j}+a_{k-1,j-1}}j\tag{12}\\ &=\frac1{j^k}\sum_{i=j-1}^{k-1}j^ia_{i,j-1}\tag{13} \end{align} $$(13)$$ follows from $$(12)$$ after unrolling the recursion.

Properties of the Coefficients

We will show that for some $$b_{j,n}$$, $$a_{k,j}=\sum_{n=1}^j(-1)^{n-1}b_{j,n}\left(\frac1n\right)^k\tag{14}$$ Since $$a_{k,1}=1$$, $$(14)$$ is true for $$j=1$$ with $$b_{1,1}=1$$ and $$b_{1,n}=0$$ for $$n\gt1$$.

Suppose that $$(14)$$ is true for $$j-1$$. Applying $$(13)$$ to $$(14)$$ yields \begin{align} a_{k,j} &=\frac1{j^k}\sum_{i=j-1}^{k-1}j^ia_{i,j-1}\\ &=\frac1{j^k}\sum_{i=j-1}^{k-1}\sum_{n=1}^{j-1}(-1)^{n-1}b_{j-1,n}\left(\frac jn\right)^i\\ &=\frac1{j^k}\sum_{n=1}^{j-1}(-1)^{n-1}b_{j-1,n}\frac{\left(\frac jn\right)^k-\left(\frac jn\right)^{j-1}}{\frac jn-1}\\ &=\sum_{n=1}^{j-1}(-1)^{n-1}\underbrace{\,\frac{nb_{j-1,n}}{j-n}\vphantom{\sum_{n=1}^{j-1}}\,}_{b_{j,n}}\left(\frac1n\right)^k\underbrace{-\sum_{n=1}^{j-1}(-1)^{n-1}\frac{nb_{j-1,n}}{j-n}\left(\frac jn\right)^{j-1}}_{(-1)^{j-1}b_{j,j}}\left(\frac1j\right)^k\tag{15} \end{align} which is of the form of $$(14)$$ with $$b_{j,n}=\frac{nb_{j-1,n}}{j-n}\tag{16}$$ for $$n\lt j$$ and $$b_{j,j}=-\sum_{n=1}^{j-1}(-1)^{j-n}b_{j,n}\left(\frac jn\right)^{j-1}\tag{17}$$ Thus, $$(14)$$ is true for $$j$$.

Computing $$\boldsymbol{b_{j,n}}$$

Equation $$(14)$$, $$(16)$$, and $$(17)$$ allow us to give formulas for $$a_{k,j}$$ for each $$j$$: \begin{align} a_{k,1}&=1\vphantom{\left(\frac11\right)^k}\\ a_{k,2}&=1-2\left(\frac12\right)^k\\ a_{k,3}&=\frac12-4\left(\frac12\right)^k+\frac92\left(\frac13\right)^k\\ a_{k,4}&=\frac16-4\left(\frac12\right)^k+\frac{27}2\left(\frac13\right)^k-\frac{32}3\left(\frac14\right)^k\\ a_{k,5}&=\frac1{24}-\frac83\left(\frac12\right)^k+\frac{81}4\left(\frac13\right)^k-\frac{128}3\left(\frac14\right)^k+\frac{625}{24}\left(\frac15\right)^k\\ a_{k,6}&=\frac1{120}{-}\frac43\left(\frac12\right)^k{+}\frac{81}4\left(\frac13\right)^k{-}\frac{256}3\left(\frac14\right)^k{+}\frac{3125}{24}\left(\frac15\right)^k{-}\frac{324}5\left(\frac16\right)^k \end{align}\tag{18} Looking at $$b_{j,j}$$ in $$(18)$$, a good guess appears to be $$b_{j,j}=\frac{j^j}{j!}\tag{19}$$ Combining $$(16)$$ and $$(19)$$, we get $$b_{j,n}=\binom{j}{n}\frac{n^j}{j!}\tag{20}$$ which satisfies $$(16)$$ and $$(17)$$, validating the guess made for $$(19)$$.

Computing $$\boldsymbol{a_{k,j}}$$

Putting together $$(14)$$ and $$(20)$$ gives $$a_{k,j}=\sum_{n=1}^j(-1)^{n-1}\binom{j}{n}\frac{n^j}{j!}\left(\frac1n\right)^k\tag{21}$$ Note that for $$j\gt k$$, the sum in $$(21)$$ is an order $$j$$ difference of a degree $$j-k$$ polynomial, hence $$a_{k,j}=0$$, which shows that $$(21)$$ is valid even for $$j\gt k$$.

Simpler Formula for the Sums

Applying $$(9)$$ and $$(11)$$ to $$(21)$$ yields \begin{align} \sum_{n=1}^\infty\frac{n^{n-k}}{e^nn!} &=\sum_{j=1}^k(-1)^{j-1}a_{k,j}\\ &=\bbox[5px,border:2px solid #C0A000]{\sum_{j=1}^k\sum_{n=1}^j(-1)^{j-n}\binom{j}{n}\frac{n^j}{j!}\left(\frac1n\right)^k}\tag{22} \end{align} For each $$k$$, the sum in $$(22)$$ has $$\frac{k^2+k}2$$ terms. This is as close to a closed formula for the sums as I have found.

Verification of the Formula

As noted after equation $$(21)$$, the inner sum in $$(22)$$ for $$j\gt k$$ is $$0$$. That is, \begin{align} \sum_{j=1}^k\sum_{n=1}^j(-1)^{j-n}\binom{j}{n}\frac{n^j}{j!}\left(\frac1n\right)^k &=\sum_{j=1}^\infty\sum_{n=1}^j(-1)^{j-n}\binom{j}{n}\frac{n^j}{j!}\left(\frac1n\right)^k\tag{23}\\ &=\sum_{n=1}^\infty\sum_{j=n}^\infty(-1)^{j-n}\frac1{n!}\frac{n^{j-k}}{(j-n)!}\tag{24}\\ &=\sum_{n=1}^\infty\sum_{j=0}^\infty(-1)^j\frac1{n!}\frac{n^{j+n-k}}{j!}\tag{25}\\ &=\sum_{n=1}^\infty\frac{n^{n-k}}{e^nn!}\tag{26} \end{align} Explanation
$$(23)$$: extend the sum in $$j$$ since the inner sums vanish for $$j\gt k$$
$$(24)$$: switch order of summation and simplify the summand
$$(25)$$: substitute $$j\mapsto j+n$$
$$(26)$$: evaluate the sum in $$j$$

• @ViktorGlombik: using the recursion for $P_k$, the polynomials tend to $xe^{-x}$. Thus, the coefficient for $x^{n+1}$ tends to $\frac{(-1)^n}{n!}$.
– robjohn
Sep 6, 2019 at 20:20
• Those coefficients only tend to $\frac{(-1)^n}{n!}$, they are not equal to it. Note that in $P_k(x)$, the coefficient of $x^k$ is $\frac{(-1)^{k-1}}{k!}$, not $\frac{(-1)^{k-1}}{(k-1)!}$.
– robjohn
Sep 6, 2019 at 20:37
• (+1) Nice answer. Although I see similarities in both answers, this one is quite friendly and easy-to-read. Anyway, do you have any guess on some properties of $P_k$? Given that $u_k$'s admit combinatoric interpretation using labelled trees, I suspect that $P_k$'s may possibly have been studied... Sep 6, 2019 at 20:52
• @SangchulLee: The only thing I know so far is that $P_k(x)\to xe^{-x}$ by equation $(7)$ above. I am looking for a pattern for the coefficients (the denominators are quite orderly, but the numerators are stickier).
– robjohn
Sep 6, 2019 at 21:49
• @ViktorGlombik: equations $(18)$ shows exactly how $a_{k,j}\to\frac1{(j-1)!}$.
– robjohn
Sep 8, 2019 at 11:04

For each $$k \geq 1$$ we define $$T_k$$ by

$$T_k(z) := \sum_{n\geq 1} \frac{n^{n-k}}{n!}z^n.$$

We make several observations.

1. $$T_1(z) = -W(-z)$$ from the Taylor series of the Lambert $$W$$-function, which is

$$W(z) = \sum_{n \geq 1} (-n)^{n-1} \frac{z^n}{n!}.$$

In particular, we get

$$T_1(we^{-w}) = w$$

for all $$w \leq 1$$. As a special case, we get $$T_1(e^{-1}) = 1$$.

2. $$T_k'(z) = T_{k-1}(z)/z$$. In particular,

$$\frac{\mathrm{d}}{\mathrm{d}w} T_k(w e^{-w}) = \frac{1-w}{w}T_{k-1}(w e^{-w}).$$

From this, we can compute $$T_2$$ as follows:

\begin{align*} T_2(we^{-w}) = \int \frac{1-w}{w}T_{1}(w e^{-w}) \, \mathrm{d}w = w - \frac{w^2}{2}. \end{align*}

In particular,

$$\sum_{n\geq 1} \frac{n^{n-2}}{n! e^n} = T_2(e^{-1}) = \frac{1}{2}.$$

A similar argument shows that each $$T_k(we^{-w})$$ is a degree $$k$$ polynomial over $$\mathbb{Q}$$ without constant term, which can be computed recursively in $$k$$. For instance, we have

$$\begin{array}{|c|ccccccc|} \hline k & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline T_k(e^{-1}) & 1 & \dfrac{1}{2} & \dfrac{5}{12} & \dfrac{7}{18} & \dfrac{1631}{4320} & \dfrac{96547}{259200} & \dfrac{40291823}{108864000} \\ \hline \end{array}$$

Addendum - closed form of $$P_n$$.

As above, let $$P_n$$ be the polynomial such that $$T_n(z) = P_n(T_1(z))$$. Then we claim that the following generating function identity holds.

$$\sum_{n=1}^{\infty} P_n(z)t^n = - \sum_{n=1}^{\infty} \prod_{i=1}^{n} \frac{tz}{t-i}. \tag{*}$$

Once we prove this, we can find the coefficient of $$z^k$$ in $$P_n(z)$$ as

$$[z^k] P_n(z) = - [t^n] \frac{t^k}{(t-1)(t-2)\cdots(t-k)} = \frac{1}{k!} \sum_{j=1}^{k} (-1)^{k-j} \binom{k}{j} \frac{1}{j^{n-k}}.$$

Consequently, we get

$$P_n(z) = \sum_{k=1}^{n} \Bigg( \sum_{j=1}^{k} (-1)^{k-j} \binom{k}{j} \frac{1}{j^{n-k}} \Bigg) \frac{z^k}{k!}.$$

We conclude with the proof of $$\text{(*)}$$.

Proof of $$\text{(*)}$$. Using partial fraction decomposition, we find that

\begin{align*} - \frac{t^k}{(t-1)\cdots(t-k)} &= -1 - \sum_{j=1}^{k} (-1)^{k-j} \frac{j^k}{(j-1)!(k-j)!} \frac{1}{t-j} \\ &= -1 + \sum_{j=1}^{k} \frac{(-1)^{k-j}}{j!(k-j)!} j^k \sum_{n=0}^{\infty} \frac{t^n}{j^n}. \end{align*}

Multiplying $$z^k$$ and summing over $$k = 1, 2, \cdots$$, we get

\begin{align*} - \sum_{k=1}^{\infty} \frac{t^k z^k}{(t-1)\cdots(t-k)} &= -\frac{z}{1-z} + \sum_{k=1}^{\infty} z^k \sum_{j=1}^{k} \frac{(-1)^{k-j}}{j!(k-j)!} j^k \sum_{n=0}^{\infty} \frac{t^n}{j^n} \\ &= -\frac{z}{1-z} + \sum_{n=0}^{\infty} \sum_{j=1}^{\infty} \frac{t^n}{j! j^n} \sum_{k=j}^{\infty} z^k \frac{(-1)^{k-j}}{(k-j)!} j^k \\ \\ &= -\frac{z}{1-z} + \sum_{n=0}^{\infty} \sum_{j=1}^{\infty} \frac{t^n}{j!} j^{j-n} (ze^{-z})^j \\ &= -\frac{z}{1-z} + \sum_{n=0}^{\infty} T_n(ze^{-z}) t^n. \end{align*}

Now the desired equality follows by noting that

$$T_0(ze^{-z}) = \frac{z}{1-z} \qquad \text{and} \qquad T_n(ze^{-z}) = P_n(z)$$

for $$n \geq 1$$.

• @ViktorGlombik, We already know that $T_1$ is the inverse of $w\mapsto we^{-w}$, and so, $T_2(z)=T_1(z)-\frac{T_2(z)^2}{2}$ and so forth. Sep 5, 2019 at 17:11
• I see. Thank you. You surely mean $T_1^2$ in the last term, right? Sep 5, 2019 at 17:11
• This OEIS entry seems to be the required function. Sep 6, 2019 at 4:13
• @ViktorGlombik, No. what is true is that $P_k(w)$ converges to $we^{-w}$ as $k\to\infty$ (at least pointwise and suspectedly locally uniformly), but $P_k$'s for $k \geq 2$ are never a Taylor polynomial of $we^{-w}$. This is easily seen from the table $\text{(8)}$ of values of $P_k$'s provided in robjohn's answer. Sep 6, 2019 at 20:41
• @SangchulLee Right, I see it now. Sep 6, 2019 at 20:51