# Thread: Sum = Integral

1. ## Sum = Integral

I have to prove that $\sum_{x=0}^{k-1}\frac{(\lambda w)^{x}e^{- \lambda w}}{x!}=\int_{\lambda w}^{\infty }\frac{z^{k-1}e^{-z}}{(k-1)!}dz$

How would I go about doing that?

2. ## Re: Sum = Integral

Originally Posted by BrownianMan
I have to prove that $\sum_{x=0}^{k-1}\frac{(\lambda w)^{x}e^{- \lambda w}}{x!}=\int_{\lambda w}^{\infty }\frac{z^{k-1}e^{-z}}{(k-1)!}dz$

How would I go about doing that?
Induction on k

CB

3. ## Re: Sum = Integral

How do I start?

4. ## Re: Sum = Integral

Originally Posted by BrownianMan
How do I start?
You prove a base case, for this show that the formula is true when $k=1$ that is show that:

$\frac{(\lambda w)^0e^{-\lambda w}}{0!}=e^{-\lambda w}=\int_{\lambda w}^{\infty} \frac{z^0 e^{-z}}{0!}\; dz=\int_{\lambda w}^{\infty} e^{-z}\; dz$

That is show:

$e^{-\lambda w}=\int_{\lambda w}^{\infty} e^{-z}\; dz$.

Next assume it true for some $k$, then prove it true for $k+1$. This will involve integration by parts, start by using the induction assumption:

$\sum_{x=0}^{(k+1)-1}\frac{(\lambda w)^xe^{-\lambda w}}{x!}=\sum_{x=0}^{k-1}\frac{(\lambda w)^xe^{-\lambda w}}{x!}+\frac{(\lambda w)^ke^{-\lambda w}}{k!}\\ \\ \\ \phantom{xxxxxxxxx} =\int_{\lambda w}^{\infty} \frac{z^{k-1} e^{-z}}{(k-1)!}\; dz+\frac{(\lambda w)^ke^{-\lambda w}}{k!}$

Now show using integration by parts on the integral on the right that:

$\int_{\lambda w}^{\infty} \frac{z^{k-1} e^{-z}}{(k-1)!}\; dz+\frac{(\lambda w)^ke^{-\lambda w}}{k!}=\int_{\lambda w}^{\infty} \frac{z^{(k+1)-1} e^{-z}}{((k+1)-1)!}\; dz$

CB

5. ## Re: Sum = Integral

Thanks so much!