I'm trying to differentiate the following

$F_{Y_K} = 1 - \sum\limits_{n = 0}^{k-1} \frac{{(\lambda y)}^n e^{-\lambda y}}{n!}$

This turns out to be

$\frac{d}{dy} F_{Y_K} = \frac{{\lambda^k y^{k-1} e^{-\lambda y}}}{(k-1)!}$

I have tried to differentiate, but don't know what to do with the summation.

This is what I have:

$\frac{d}{dy} F_{Y_K} = - \sum\limits_{n = 0}^{k-1} \frac{\lambda^n}{n!}\frac{d}{dy} (y^n e^{-\lambda y}) = -\sum\limits_{n = 0}^{k-1} \frac{\lambda^n}{n!} [y^n(-\lambda e^{-\lambda y}) + (n y^{n-1})e^{-\lambda y}] = -\sum\limits_{n = 0}^{k-1} \frac{\lambda^n}{n!}e^{-\lambda y}(n y^{n-1} - \lambda y^n)$

How do I continue?

  • 2
    $\begingroup$ Imagine $k$ large, like $12$. You have a difference of sums. Note that for example the term in the first sum corresponding to $n=4$ cancels the term in the second sum corresponding to $n=3$. And so on for almost all terms. $\endgroup$ – André Nicolas Oct 7 '15 at 17:10
  • $\begingroup$ Thanks! I got it, and also realized I forgot the negative sign in my derivative. $\endgroup$ – Rayne Oct 7 '15 at 17:52
  • 1
    $\begingroup$ You are welcome. A formal write up can be done by separating the two sums. In the one that has the $n$ on top, cancel with the $n$ in $n!$, getting $\frac{1}{n-1}!$. The replace everywhere in that sum $n-1$ by $j$, and then $j$ by $n$. Now observe the cancellation with terms of the second sum. I did not feel like writing it up since it is tedious TeX-wise. $\endgroup$ – André Nicolas Oct 7 '15 at 18:05

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