Can the Sum of Powers Over an Integral Approach 1 as n Tends to Infinity?

[eljose]

i would like to know how to prove this equality:

\frac{1+2^p+3^p+...+n^p}{\int_0^{n}dxx^p}\rightarrow{1}

for n\rightarrow{\infty} of course p>0

i don,t know if is repeated (sorry in that case)..

 

[HallsofIvy]

Draw the graph of y= x^p. Now draw a series of rectangles, base from x= 1 to 2, height 1^p= 1, base 2 to 3, height 2^p, etc. Observe that each rectangle lies under the graph and so the total area of the rectangles (up to x= n) is less than the area under the curve. That is, 1 is an upper bound for the sequence as n goes to infinity. It should not be difficult to show that the sequence is increasing.

 

[Hurkyl]

Go look at the integral comparison test. The method of proof is exactly what you want. (And what Integral describes)

But, the sum is "well-known" (for integer exponents):

1^k + 2^k + 3^k + \cdots + n^k = \frac{n^{k+1} }{k+1} + \frac{n^k}{2} + O(n^{k-1})

 

[mathwonk]

it is easy to prove the sum ^k + ...+n^k, equals n^(k+1)/[k+1] + O(n^k), by induction.
just use the binomial theorem on (a+1)^(k+1) - a^(k+1). Expand and add up over all a=1,...,n.
you get a telescoping sum that shows (n+1)^(k+1) - 1^(k+1)
= (k+1) )(1^k +...+n^k) + sums of all lower powers of integers summed from 1 to n.
QED.
reference, p.27 of courant's calculus book vol 1, in the precalculus section.