What is the probability a computer ever finishes all jobs in its job queue (clear details inside) Problem: A computer receives $N_{i}$
  new jobs at the beginning of each day where $\mathbb{P}\left(N_{i}=k\right)=r^{k}\left(1-r\right)$
  for $i\geq1$
 . The computer already has $1$ job in the queue when $i=1$
  and it can only complete $1$ job per day. What is the probability there will ever be a day which ends with the job queue being empty?
Attempted solution: At the end of day $N\geq1$
  the total number of jobs that arrived (apart from the first job) is $M_{N}:=\sum_{i=1}^{N}N_{i}$
  and the total number of jobs that were in the queue is $M_{N}+1$
 . Since the total number of jobs completed by day $N$
  (assuming there was always at least one job in the queue) is $N$
  we know that if $M_{N}+1>N$
  or equivalenty $M_{N}>N-1$
  the queue would not be empty at the end of the $N'\text{th}$
 day. It can be seen that each $N_{i}$
  is actually geometrically distributed on $\left\{ 0,1,2,...\right\}$ 
 with parameter $q=\left(1-r\right)$
 . These $N_{i}$
  are also independent and it is known that the sum of $N$
  i.i.d $\text{Geo}\left(q\right)$
  rvs is distributed negative binomial with parameters $\left(N,1-q\right)$
 , thus we have that $M_{N}\sim\text{NB}\left(N,r\right)$
  which means that for $k\in\left\{ 0,1,2,...\right\}$ 
  we have $$\mathbb{P}\left(M_{N}=k\right)={k+r-1 \choose k}r^{k}\left(1-r\right)^{N}$$
 Thus$$\mathbb{P}\left(M_{N}>N-1\right)=1-\mathbb{P}\left(M_{N}\leq N-1\right)=1-\sum_{k=0}^{N-1}\mathbb{P}\left(M_{N}=k\right)=1-\sum_{k=0}^{N-1}{k+r-1 \choose k}r^{k}\left(1-r\right)^{N}=1-\left(1-r\right)^{N}\sum_{k=0}^{N-1}{k+r-1 \choose k}r^{k}$$
 Now I have no idea what to do with this sum. In general the probability there is ever a day where the computer finishes all tasks is the complement of the probability it never finishes which is: $$\sum_{N=1}^{\infty}\mathbb{P}\left(M_{N}>N-1\right)$$
 So assuming I knew how to calculate $\mathbb{P}\left(M_{N}>N-1\right)$
  and I could sum all these values I would have my answer... I would appreciate both feedback at my attempted solution or in case this is a dead-end an alternative more effective approach. 
 A: An equivalent question is:

Take a $1$-dimensional random walk: start at $1$; at each step, increase by $1$ with probability $r$ and decrease by $1$ with probability $1-r$. What is the probability that you ever end up at $0$?

To see this equivalence, note that the probability in this random walk that you increase exactly $k$ times before your next decrease is always $r^k(1-r)$.
If $r \leq \frac{1}{2}$, this will clearly happen with probability $1$. (If you don't find this clear when $r$ is exactly $\frac{1}{2}$, google "gambler's ruin.")
If $r>\frac{1}{2}$, let $p_n$ be the probability that we hit $0$, given that we are
currently at $n$. Then we know that:
$$
p_0=1\\
\lim_{n \to \infty} p_n=0\\
p_n=(1-r)p_{n-1}+rp_{n+1}, \quad n >0\\
$$
The third relation can be rewritten as a linear recurrence for $p_n$:
$$
p_{n+2}-\frac{1}{r}p_{n+1}+\frac{1-r}{r}p_n=0, \quad n>1
$$
The characteristic equation of this recurrence, $x^2-\frac{1}{r}x+\frac{1-r}{r}$, has roots $1$ and $\frac{1-r}{r}$, and so we must have
$$
p_n=A+B\left(\frac{1-r}{r}\right)^n
$$
for some constants $A,B$. But $\lim_{n \to \infty} p_n=0$, and so $A=0$; also, $p_0=1$, so $B=1$. So $p_n=\left(\frac{1-r}{r}\right)^n$. In particular, $p_1=\frac{1-r}{r}$; thus the probability is $\frac{1-r}{r}$ when $r > \frac{1}{2}$.
