n n+1 is of course correct, just as hhsilver has quite cleverly intuited.
No clue why mathdotcom gave him such a hard time.
But just for shits and giggles, a rather informal late night mathematical proof might read something like this:
We're looking for the E(sup(Xk)) where Xk is a vector of k realizations from the uniform distribution.
For the degenerate case of k=1, the pdf(sup(X1)) = pdf(X1) = 1. Hence:
=
1 2 * p
2 | p=0 → 1
=
1 2
So E(sup(X1)) = E(X1) =
1 2
For k=2:
Pr(sup(X2) = p ∈ [0,1] )
=
Pr( (x(1) ≤ p AND x(2) = p) OR (x(1) = p AND x(2) ≤ p) )
(this is the key step)
≈ 2*p*Δ
Hence:
= 2 3 * p3 | p=0 → 1
= 2 3
Continuing with the same logic, it's clear that for k=n:
Pr(sup(Xn) = p ∈ [0,1] )
≈ n*p
n-1*Δ
=
n n+1 * p
n+1 | p=0 → 1
=
n n+1
QED
Nicely done, hh. YOu get full credit in my book.