sum of divisors

let S(n) denote the sum of the divisors of n including 1 and itself.
let [.] denote the floor function.
then how can we prove the following result :-

S(1)+S(2)+.....+S(n)= 1*[n/1] + 2*[n/2] + .... + n*[n/n]

Re: sum of divisors

Dear Peruchin,
I do not agree with your reply that induction is straightforward. How an I supposed to use the fact,
(quote:Hint: [;(n+1)=(n+1)\lfloor\frac{n+1}{n+1} \rfloor);]).

Re: sum of divisors

verify that, if m|(N+1) then m[(N+1)/m]-m[N/m] = m; otherwise it is 0.
it follows that sum(n=1 to N+1;m[(N+1)/m]) - sum(n=1 to N;m[N+/m]) is equal to the sum of the divisors of N+1 = S[n+1]. The theorem follows.

Re: sum of divisors

Sorry for all of the previous mistakes. Let me try again.

Verify that

m[(N+1)/m] - m[N/m] = m if m divides N+1
= 0 otherwise

It follows that

sum(m=1 to N+1; m[(N+1)/m]) - sum(m=1 to N; m[N/m])

= the sum of the divisors of N+1

= S[N+1].

The theorem follows.

Re: sum of divisors

Induction over n is straightforward.
Hint: (n+1) = (n+ 1)[(n+1)/(n+1)],
where the brackets stand for the floor function, the same as the integer part function for n>0.
peruchin