I'm looking to find a closed-form expression for the sum $$S = \sum_{n=0}^N e^{-x/2} L_n^{0}(x),$$ where $L_n^{0}$ is the $n$th Laguerre polynomial. Using the formula $$L_n^{\alpha}(x) = \sum_{m=0}^n (-1)^m \binom{n+\alpha}{n-m} \frac{x^m}{m!},$$ this is equivalent to finding a closed form for $$\sum_{n=m}^N \binom{n}{m}.$$ Any help would be appreciated. An asymptotic form would also be useful for the quantity $\frac{S}{N}$.

  • $\begingroup$ Check definition for S. I did not see $k$ in the summation. $\endgroup$
    – mike
    Jun 23, 2014 at 16:24
  • 1
    $\begingroup$ $\sum_{n=m}^N \left(\begin{array}{c} n \\ m \end{array}\right)=\binom{N+1}{m+1}$ $\endgroup$ Jun 23, 2014 at 16:25
  • $\begingroup$ @mike: Fixed dependence of sum, thanks. $\endgroup$ Jun 23, 2014 at 16:32
  • $\begingroup$ @ClaudeLeibovici can you give a quick sketch of the proof? $\endgroup$ Jun 23, 2014 at 16:33

1 Answer 1


From the identity $\sum_{n=m}^N \binom{n}{m} = \binom{N+1}{m+1}$ we have that

$$S = e^{-x/2} \sum_{m=0}^N (-1)^m \binom{N+1}{m+1} \frac{x^m}{m!} = e^{-x/2}L_N^{1}(x)$$


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