What is lim n --> oo x_n = a(1^a + 2^a +.....n^a)/[n^(a +1)]?

Question

If a <=-1, then, we can show a_n --> -oo
If a >= 0, this is a sequence of Riemann sums of f(x) = x^a over [0,1] and a_n --> Integral [0 to 1] x^a dx = = (x^(a +1))/(a +1) [ o to 1] = 1/(a +1)

My question is, what happens when a is in (-1, 0)? Then, this is not a sequence of Riemann sums over [0,1] because x^a is not defined at x = 0 when a < 0. It sems the limit is still 1/(a +1), but is this true?

Answer 1

S(a,n)=a*(1^a + 2^a +.....+n^a)/[n^(a +1)]
=a/n*((1/n)^a+(2/n)^a+...+(n/n)^a)
=a*((1/n)^a+(2/n)^a+...+(n/n)^a)/n
Note that ((1/n)^a+(2/n)^a+...+(n/n)^a)/n is just the mean of the set {(1/n)^a,(2/n)^a,...,1}, so this mean must lie somewhere between the extreme values (1/n)^a and 1. This means
a*((1/n)^a+(2/n)^a+...+(n/n)^a)/n must lie between a*(1/n)^a and a. For a in (-1,0), a*(1/n)^a<=lim{n->infinity}S(a,n)<=a, so in the limit we get
lim{n->infinity}S(a,n)<=a.

Answer 2

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