A transmitter is sending data packets. But some of these packets can get lost on the way. $N_r$ is the average number of transmission attempts including the successful one. The probability for j transmission attempts is:
$P(j) = PER^{j-1} (1-PER)$ with PER = packet error rate, so a probability
The expectation value of $N_r$ is: $\sum_{j=1}^{\infty} j\ PER^{j-1} (1-PER) = \frac{1}{1-PER}$
Well, I really don't understand how to come up with the right side of the last equation.
For $|x|<1$, the geometric series gives $\sum_{j=0}^{\infty}x^{j}=\frac{1}{1-x}$. Hence, $\sum_{j=1}^{\infty}x^{j}=\frac{1}{1-x}-1=\frac{x}{1-x}$. We can differentiate this series term-by-term to obtain
$$ \sum_{j=1}^{\infty}jx^{j-1}=\frac{1}{(1-x)^{2}}. $$
Hence, $$ \sum_{j=1}^{\infty}jx^{j-1}(1-x)=\frac{1}{1-x}. $$
Now, take $x=PER$ in your example.