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I am trying to work these out: $$\sum\limits_{u=0}^{k}{\left( \begin{matrix} r+u \\ u \\ \end{matrix} \right){{\left( {\alpha} \right)}^{u}}{{\left( {1-\alpha} \right)}^{r+1}}}$$ $$\sum\limits_{u=0}^{k}{\left( \begin{matrix} r+u \\ u \\ \end{matrix} \right){{\left( \alpha \right)}^{u}}{{\left( 1-\alpha \right)}^{r+1}}\sum\limits_{m=u+r+1}^{\infty }{\frac{{{e}^{-\lambda }}{{\lambda }^{m}}}{m!}}}$$

Thanks!

Eln
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  • You can have a closed form for the sum in terms of the hypergeometric function. – Mhenni Benghorbal Feb 16 '13 at 00:49
  • What is $k$? If "sufficiently large" (so that you can live with $k \rightarrow \infty$), then $\binom{-r - 1}{u} = (-1)^u \binom{r + u}{u}$, and you have a good approximation for the first sum if $\lvert \alpha \rvert$ is small. – vonbrand Feb 16 '13 at 03:17
  • Thanks @ Mhenni Benghorbal. I found that also I can use regularized beta function too. – Eln Feb 16 '13 at 04:09
  • Thanks @ vonbrand. I can definitely use that for approximation, in some case k could be quite large. For the second integral, I was trying to change the order of summations, see if that helps. – Eln Feb 16 '13 at 04:23

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