I don't understand how did it go from left hand side to the right hand side for this equation: $$ \sum_{r=n}^{\infty}[\frac{\lambda(\lambda t)^{r-1}e^{-\lambda t}}{(r-1)!} - \frac{\lambda(\lambda t)^re^{-\lambda t}}{r!}] = \frac{\lambda(\lambda t)^{n-1}e^{-\lambda t}}{(n-1)!} $$
How I did(and got stuck): $$ \sum_{r=n}^{\infty}[\frac{\lambda(\lambda t)^{r-1}e^{-\lambda t}}{(r-1)!} - \frac{\lambda(\lambda t)^re^{-\lambda t}}{r!}] = \lambda e^{-\lambda t} \sum_{r=n}^{\infty}\frac{(\lambda t)^{r-1} (r-\lambda t)}{r!} $$
This is Example 5.2 of the book Stochastic Processes: An Introduction by P.W. Jones and P.Smith, 3rd edition.
The entire Example + Solution is as follows:
Example 5.2 Incoming telephone calls to an operator are assumed to be a Poisson process with parameter λ. Find the density function of the length of time for n calls to be received, and find the mean time and variance of the random variable of the length of time for $n$ calls.
We are now interested in the time $T_n$, which is the earliest time at which the random variable $N(t) = n$ occurs, and its distribution. The probability distribution of the random variable Tn is given by
$\begin{equation} \begin{split} F(t) &= P\{T_n≤t\} \\ &= P\{\text{n or more calls have arrived in the time interval} (0, t)\} \\ &= p_n(t)+p_{n+1}(t)+ ···\\ &= \sum^{∞}_{r=n} p_r(t) \\ &= \sum^{∞}_{r=n} \frac{(λt)^r e^{−λt}}{r!} \end{split} \end{equation}$
using (5.1). The corresponding density function is
$$ f(t)=\frac{dF(t)}{dt}=\sum_{r=n}^{\infty}[\frac{\lambda(\lambda t)^{r-1}e^{-\lambda t}}{(r-1)!} - \frac{\lambda(\lambda t)^re^{-\lambda t}}{r!}] = \frac{\lambda(\lambda t)^{n-1}e^{-\lambda t}}{(n-1)!} $$
Thank you in advance!
Rewrite the sum as
$$ \sum_{r=n}^{\infty}[\frac{\lambda(\lambda t)^{r-1}e^{-\lambda t}}{(r-1)!} - \frac{\lambda(\lambda t)^re^{-\lambda t}}{r!}] = \sum_{r=n}^{\infty}\frac{\lambda(\lambda t)^{r-1}e^{-\lambda t}}{(r-1)!} - \sum_{r=n}^{\infty}\frac{\lambda(\lambda t)^re^{-\lambda t}}{r!}$$
(it is absolutely convergent, so the rearrangement is legal).
Observe that the first sum is
$$ \sum_{r=n}^{\infty}\frac{\lambda(\lambda t)^{r-1}e^{-\lambda t}}{(r-1)!} = \sum_{r=n-1}^{\infty}\frac{\lambda(\lambda t)^{r}e^{-\lambda t}}{(r)!} = \frac{\lambda(\lambda t)^{n-1}e^{-\lambda t}}{(n-1)!} + \sum_{r=n}^{\infty}\frac{\lambda(\lambda t)^{r}e^{-\lambda t}}{(r)!} $$
So both infinite summations are identical and cancel each out.