Sums-Of-Powers over Minimum Divisors

[Max]

Leroy Quet wrote:

> Let s(r,m) =
> 
> ---
> \      r
>  >    k
> /
> ---
> k|m
> 1<= k <= sqrt(m)
> 
> (which is, in linear-mode)
> 
> sum{k|m,1<= k<= sqrt(m)}   k^r.
>
> So, we have s(r,m) is 
> the sum of the r-powers taken over the lower half of the positive 
> divisors of m.
> 
> For example, s(1,m) is:
> http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?A
> num=A066839
>  
> 
>  If r is > 0 (r = any *positive* real), then:
> 
> 
> limit{m -> oo} 
>            m
>           ---
>   1       \
> -------    >   s(2r,k)       =
> m^(r+1)   /
>           ---
>           k=1
> 
> 
>     1
> ----------    (?)
> 2 r (r+1)
> 
> 
> Linear-mode:
> 
> limit{m->oo} 
>       (1/m^(r+1)) sum{k=1 to m} s(2r,k)  = 
> 
> 1/(2 r (r+1))     (?)

Each number q<=sqrt(m) is a divider of exactly [m/q]-q numbers >=q^2.
Hence,

sum{k=1 to m} s(2r,k) = sum{1<=q<=sqrt(m)} ([m/q]-q)*q^(2r) =
sum{1<=q<=sqrt(m)} [m/q]*q^(2r) - q^(2r+1) =
m*sqrt(m)^(2r)/(2r) - sqrt(m)^(2r+2)/(2r+2) + O(sqrt(m)^(2r+1)) =
m^(r+1)/(2*r*(r+1)) + O(m^(r+1/2))

implying the limit above.

Regards,
Max