How do you sum this series? $$\sum _{y=1}^m \frac{y}{(m-y)!(m+y)!}$$

My attempt:

$$\frac{y}{(m-y)!(m+y)!}=\frac{y}{(2m)!}{2m\choose m+y}$$

My thoughts were, sum this from zero, get a trivial answer, take away the first term. But actually I don't think this will work very well.

This question was originally under probability, but the problem is that I can't sum a series and really has nothing to do with probability (reason for the first comment)

  • $\begingroup$ What are positive paths? I can't identify any common features of the paths in the image. $\endgroup$
    – joriki
    May 2, 2013 at 10:24
  • $\begingroup$ Each move we get $S_k=S_{k-1}\pm 1$, I want $\min_{1\leq k \leq n} S_k>0$, so all of the lines which stay strictly above the x-axis. $\endgroup$
    – shilov
    May 2, 2013 at 10:32

2 Answers 2


For example, one can write

\begin{align} \sum_{y=0}^m\frac{y}{(m-y)!(m+y)!} &= \sum_{k=0}^m\frac{m-k}{k!(2m-k)!} \\ &=\frac{m}{(2m)!}\sum_{k=0}^m{2m \choose k}-\frac{1}{(2m-1)!}\sum_{k=1}^{m}{2m-1\choose k-1} \\ &= \frac{m}{2(2m)!}\left[{2m\choose m}+\sum_{k=0}^{2m}{2m \choose k}\right]-\frac{1}{(2m-1)!}\sum_{k=0}^{m-1}{2m-1\choose k} \\ &= \frac{m}{2(2m)!}\left[{2m\choose m}+\left(1+1\right)^{2m}\right]-\frac{1}{2(2m-1)!}\sum_{k=0}^{2m-1}{2m-1\choose k} \\ &= \frac{m}{2(2m)!}{2m\choose m}+\frac{m\cdot 2^{2m}}{2(2m)!}-\frac{2^{2m-1}}{2(2m-1)!} \\ &= \frac{m}{2(2m)!}{2m\choose m}\;. \end{align} All we have used in the way is that $\displaystyle{n\choose k} ={n\choose n-k}$ and that $\displaystyle(1+1)^n=\sum_{k=0}^n{n\choose k}$.

  • $\begingroup$ I took the liberty of arranging the equations in somewhat more orderly fashion :-) $\endgroup$
    – joriki
    May 5, 2013 at 20:34
  • $\begingroup$ Could you clarify how you obtained the first equality? $\endgroup$ May 5, 2013 at 20:39
  • 1
    $\begingroup$ @DepeHb: The first equality comes from a change of summation index: $y=m-k$. $\endgroup$ May 7, 2013 at 21:08

Try maxima on this. Load zeilberger, and:

GosperSum(y * binomial(2 m, m + y), y, 1, m)

gives $$\dfrac{1}{2}(m + 1) \binom{2 m}{m + 1}$$

  • $\begingroup$ Hi, I used Mathematica to sum it, but I want to know how the technique $\endgroup$
    – shilov
    May 3, 2013 at 16:13

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