Harmonic Number Sum Congruence

[Leroy Quet]

[update, also posted to sci.math]

qqquet at mindspring.com (Leroy Quet) wrote in message 
news:<b4be2fdf.0401091745.5cb677f8 at posting.google.com>...
> qqquet at mindspring.com (Leroy Quet) wrote in message 
news:<b4be2fdf.0401081659.60505a7f at posting.google.com>...
> > Let H(n) = sum{k=1 to n} 1/k, 
> > the n_th harmonic number.
> > 
> > Prove, for each m = positive integer,

>.... 
 
> 
> We can rewrite the above so it becomes, perhaps to some reading this,
> more simply stated and more natural.
> 
> (Take sum from 1 to m, instead of from 1 to m-1)
> 
> 
> sum{k=1 to m}  H(k) k! (m-k)!
> 
> 
> 
> is congruent to
> 
> 
> 
> m! *h(m)   (mod(m+1)),
> 
> 
> where h(m) = sum{k=0 to floor((m-1)/2)}  1/(m -2k)  .
> 
> 
> 
> Now, h(m) =
> sum{k=0 to floor((m-1)/2)}  1/(m -2k)
> is interesting in itself, for it is (analogously to
> "double-factorials") the sum of the reciprocal of EVERY-OTHER positive
> integer <=m
> (even or odd, depending on m).
> 
> 
>.... 

 
Actually this is easily proved (in a way I did not prove it originally).
 
Since (m-k)! is congruent to 
m!/k! (-1)^(m+k)  (mod{m+1}), 


sum{k=1 to m}  H(k) k! (m-k)!

is congruent to 
m! sum{k=1 to m}  H(k) (-1)^(k+m)  (mod{m+1}).

And this last sum is just equal to m!*h(m).

- 
So, generally, if {a(k)} is such that

m!*(sum{k=1 to m} a(k))

is always an integer, then

sum{k=1 to m} (sum{j=1 to k} a(j)) k! (m-k)!

is congruent to

m! sum{k=0 to floor((m-1)/2)} a(m -2k)
(mod {1+m}).

;) 
 
thanks,
Leroy Quet