1

Is there a closed form for

$$\sum_{j = 0}^k \frac{j}{(k-j)!}$$

for $k \in \mathbb{N}$.

  • The generating function is $-\frac{e^x}{(1 - x)^2}$. Maybe there's some way to extract the coefficients. – Qudit Nov 01 '17 at 07:58
  • What do you mean by generating function ? Is is that these are the coefficients of the power series corresponding to the function you gave? –  Nov 01 '17 at 08:00
  • Yes. If you can find a nice expression for the $k^{\mathrm{th}}$ derivative of the generating function divided by $k!$ that would be your closed form. Not sure if that approach will work or not though. Sometimes extracting the coefficients is trickier, but it might work. – Qudit Nov 01 '17 at 08:02
  • Are you sure about the function? I think it should be $\frac{xe^x}{(1-x)^2}$. Or am I missing something? –  Nov 01 '17 at 08:03
  • Actually, $\frac{x e^x}{(1 - x)^2}$. I dropped the $x$ I multiplied it by and added a minus sign in my original comment. – Qudit Nov 01 '17 at 08:05

3 Answers3

3

I did not find any closed form.

However, computing $$S_k=\sum_{j = 0}^k \frac{j}{(k-j)!}$$ for $(0 \leq k \leq 1000)$, it seems to be very close to $e(k-1)$. $$\left( \begin{array}{cccc} k & S_k & S_k \approx & e(k-1) \\ 5 & \frac{87}{8} & 10.87500000 & 10.87312731 \\ 10 & \frac{8877691}{362880} & 24.46453649 & 24.46453646 \\ 15 & \frac{3317652307271}{87178291200} & 38.05594560 & 38.05594560 \\ 20 & \frac{42739099682215483}{827517689856000} & 51.64735474 & 51.64735474 \end{array} \right)$$

Claude Leibovici
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3

Reversing the order of the sum (i.e. replacing $j$ with $k-j$ everywhere) gives $$ S_k = \sum_{j=0}^k \frac{j}{(k-j)!} = \sum_{j=0}^k \frac{k-j}{j!} = k\sum_{j=0}^k \frac{1}{j!} - \sum_{j=1}^k \frac{1}{(j-1)!}. $$ Now, using that $\sum_{j=0}^k \frac{1}{j!} = e \frac{\Gamma(k+1,1)}{\Gamma(k+1)}$ we recover $$ S_k = e \frac{\Gamma(k+1,1) - \Gamma(k,1)}{\Gamma(k)}. $$

ekkilop
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1

Maple gives ${\frac {1+ e(k-1) \Gamma \left( k,1 \right) }{\Gamma \left( k \right) }} $

Leox
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