I have no idea how to prove this. Can someone help?
Given a real number x, denote by [x] the largest integer less than or equal to x. Prove that the function defined
= is Riemann integrable on any interval [a,b]. Then compute
for n = 0 the term in your series is 0. so i'll assume that n > 0. let
each is integrable on any interval [a, b], because it's continuous almost everywhere on the
interval. let so is integrable on [a, b] for all n.
since by Weierstrass test, is uniformly convergent. so
is integrable on [a, b] and if a = 0 and b = 1, then this gives us:
I will do the special case . Define a sequence of functions as . Notice that . Since it follows by the Weierstrass test that the series of functions converges uniformly to a function . Each function is discontinous at for . This implies that is integrable and . It remains to compute . The graph of consists of a congruent right-angles triangles each having height and with base width . Thus, . I might have missed maybe on the exponent, but you get the idea. Now this is a geometric series which sums easily.
EDIT: Got beaten in the response by a fellow algebraist.