I have to prove that $\displaystyle \sum_{x=0}^{k-1}\frac{(\lambda w)^{x}e^{- \lambda w}}{x!}=\int_{\lambda w}^{\infty }\frac{z^{k-1}e^{-z}}{(k-1)!}dz$
How would I go about doing that?
You prove a base case, for this show that the formula is true when $\displaystyle k=1$ that is show that:
$\displaystyle \frac{(\lambda w)^0e^{-\lambda w}}{0!}=e^{-\lambda w}=\int_{\lambda w}^{\infty} \frac{z^0 e^{-z}}{0!}\; dz=\int_{\lambda w}^{\infty} e^{-z}\; dz$
That is show:
$\displaystyle e^{-\lambda w}=\int_{\lambda w}^{\infty} e^{-z}\; dz$.
Next assume it true for some $\displaystyle k$, then prove it true for $\displaystyle k+1$. This will involve integration by parts, start by using the induction assumption:
$\displaystyle \sum_{x=0}^{(k+1)-1}\frac{(\lambda w)^xe^{-\lambda w}}{x!}=\sum_{x=0}^{k-1}\frac{(\lambda w)^xe^{-\lambda w}}{x!}+\frac{(\lambda w)^ke^{-\lambda w}}{k!}\\ \\ \\ \phantom{xxxxxxxxx} =\int_{\lambda w}^{\infty} \frac{z^{k-1} e^{-z}}{(k-1)!}\; dz+\frac{(\lambda w)^ke^{-\lambda w}}{k!}$
Now show using integration by parts on the integral on the right that:
$\displaystyle \int_{\lambda w}^{\infty} \frac{z^{k-1} e^{-z}}{(k-1)!}\; dz+\frac{(\lambda w)^ke^{-\lambda w}}{k!}=\int_{\lambda w}^{\infty} \frac{z^{(k+1)-1} e^{-z}}{((k+1)-1)!}\; dz$
CB