Probability function on $\mathbb N$ - no convergence to $1$? 
Consider a box containing one red ball and one black ball. If we draw
  a black ball, we put it back and add another black ball. If we draw
  the red ball, the experiment is over. 
What is the probability $p_n$ that the red ball is drawn in the $n$-th
  drawing? Show that it's a probability function.

My thoughts:
In the first drawing, there are only two balls (red, black). So the probability is
$$p_1 = \frac{1}{2}$$
In the second drawing, if we didn't draw the red ball yet, the probability would be
$$p_2 = \frac{1}{3}\left( 1 - \frac{1}{2} \right)$$
because there are three balls now, and we multiply the probability of drawing the red ball with the counter probability $p_1^c = 1 - p_1$ from the first step.
This procedure leads to:
\begin{align}
p_3 & = \frac{1}{4} - \frac{1}{24} \\[4pt]
p_4 & = \frac{1}{5} - \frac{5}{120} \\
&\vdots \\
p_n & = \frac{1}{n+1} - \frac{1}{(n+1)!}
\end{align}
But this doesn't seem to be a probability function on $\mathbb N$:
$$\sum_{n \in \mathbb N} p_n = \sum_{n \in \mathbb N} \frac{1}{n+1} - \frac{1}{(n+1)!} = \sum_{n \in \mathbb N} \frac{n! - 1}{(n+1)!} = \infty$$
Can you help me find the mistake?
 A: Your problem is that it should be:
$$p_{n} = \frac{1}{n+1}\left(1-(p_1+p_2+\cdots+p_{n-1})\right)$$
This is the sort of problem where you should pretend to game continues forever, and instead ask what is the probability that the first red ball is chosen at turn $n$ (when selected from a bag with $n+1$ balls.)
The odds that the first $n-1$ turns were all black is:
$$\frac{1}{2}\frac{2}{3}\cdots\frac{n-1}{n}=\frac{1}{n}$$
The odds that the $n$th turn is red is $\frac{1}{n+1}$.
So the value is:$$p_n=\frac{1}{n}\frac{1}{n+1} = \frac{1}{n}-\frac{1}{n+1}$$
A: \begin{align}
p_n & = \left( 1 - \frac 1 2 \right)\left( 1 - \frac 1 3 \right)\left( 1 - \frac 1 4 \right)\left( 1 - \frac 1 5 \right) \cdots\left( 1 - \frac 1 n \right) \frac 1 {n+1} \\[10pt]
& = \left(\frac 1 2 \cdot\frac 2 3 \cdot \frac 3 4 \cdot\frac 4 5 \cdots \frac{n-1} n\right)\cdot \frac 1 {n+1} \\[10pt]
& =\frac 1 {n(n+1)} = \frac 1 n - \frac 1 {n+1}
\end{align}
These add up to $1$ because the sum telescopes:
\begin{align}
\left(1 - \frac 1 2 \right) + \left(\frac 1 2 - \frac 1 3 \right) + \left( \frac 1 3 - \frac 1 4 \right) + \cdots + \left( \frac 1 n - \frac 1 {n+1} \right) = 1 - \frac 1 {n+1} \to 1\text{ as } n\to\infty.
\end{align}
