Split into its integer part and fractional part, . There is exactly one with
By subtracting the same integer from inside the floor operations on the left and right sides of this inequality, it may be rewritten as
Therefore,
and multiplying both sides by gives
Now if the summation from Hermite's identity is split into two parts at index , it becomes
Alternate Proof[]
Consider the function
Then the identity is clearly equivalent to the statement for all real . But then we find,
Where in the last equality we use the fact that for all integers . But then has period . It then suffices to prove that for all . But in this case, the integral part of each summand in is equal to 0. We deduce that the function is indeed 0 for all real inputs .
References[]
^Savchev, Svetoslav; Andreescu, Titu (2003), "12 Hermite's Identity", Mathematical Miniatures, New Mathematical Library, 43, Mathematical Association of America, pp. 41–44, ISBN9780883856451.