# Integral to sum

## Homework Statement

Show that

$$\int_{0}^{t}\frac{\lambda^{k}u^{k-1}e^{-\lambda u}}{(k-1)!}\ du$$

is

$$\sum_{j=k}^{\infty}\frac{(\lambda\cdot t)^{j}e^{-\lambda t}}{j!}$$

Hint: Use a taylor series to express e^(-lambda*u)

## The Attempt at a Solution

I used the taylor series to rewrite the e^ term and then I interchanges the integral sign with the summation and integrated the term which leaves me with

$$\sum_{j=0}^{\infty}\left[\frac{(\lambda\cdot t)^{j}}{j!}\cdot\frac{(\lambda\cdot t)^{k}}{(k+j)(k-1)!}\right]$$

However, I have no idea how to rewrite the left-erm to e^{-\lambda t} especially since we don't sum over k (I suspect it has something to do with shifting the index to j=k but I don't really know how that can be done without creating a second summation)

The Taylor series of $e^{-\lambda u}$ will be an alternating series. Your result is not an alternating series.