With brutal force.
For k=1,..., if x is in [N/(k+1), N/k), then
the expectation of the change is \(\int_{N/(k+1)}^{N/k} (N-kx) \frac{1}{N}dx =\frac{N}{2} \frac{1}{ k (k+1)^2}\)
Then adding them altogether, we have
\(\sum_{k=1}^\infty \frac{N}{2} \frac{1}{ k (k+1)^2}=\frac{N}{2} \sum_{k=1}^\infty \{\frac{1}{ k (k+1)} - \frac{1}{ (k+1)^2}\} \)
Ignoring the N/2 for the moment, the sum of the first term is 1 and the sum of the second term (ignoring the subtraction sign) is \(\pi^2/6-1\)
Putting these two pieces together,
we have \(\frac{N}{2} \{1 -(\pi^2/6-1)\}=N(1-\pi^2/12)\)
As other people have mentioned, for this problem, brutal force is quite easy. I'm not sure if you have a cute solution other than the brutal force.
correct.
it is
\(\sum_{k=1}^\infty \large(\frac{1}{k} - \frac{1}{k+1}\right)\cdot \large(N\cdot\large(\frac{1}{k} - \frac{1}{k+1}\right)\cdot \frac{k}{2}\right)\)
one distinguishes cases depending on the number of articles bought (Pr in the first bracket) and computes the expected change in each case (the second bracket).
