Re: Accidental limit

From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 06/02/04 

Date: Wed, 02 Jun 2004 10:58:13 -0500

On Wed, 2 Jun 2004 15:36:01 +0100, "The Last Danish Pastry"
<TheLastDanishPastry@yahoo.com> wrote:

>"David C. Ullrich" <ullrich@math.okstate.edu> wrote in message
>news:o98mb0hpohjippit0krlrqt4noh6uteelv@4ax.com...
>
>> On Mon, 31 May 2004 11:46:43 +0200, "Néstor" <c@c.com> wrote:
>>
>> >Hi!
>> >
>> >I´ve just found this limit accidentally and empirically:
>> >
>> >Lim (n->oo) ( [ SUM ( i = (n^2) -> (n+1)^2 ) SQRT(i) ] ) = 1/6
>> >
>> >... where [x] means 'decimal part of x'. i.e. [3.1234]= 0.1234
>> >
>> >Is this true?
>>
>> Yes, it's true. My first guess was that something like this
>> couldn't possibly be true, but it is:
>>
>> First "note" that if N < M and N, M are integers then
>>
>> |sum_N^M f(i) - int_{N-1/2}^{M-1/2} f(x) dx|
>>
>> <= c int_{N-1/2}^{M+1/2} |f''(x)| dx.
>>
>> Using the notation o(1) for "something that tends to zero"
>> it follows that
>>
>> sum_{n^2}^{(n+1)^2} sqrt(i)
>>
>> = o(1) + int_{n^2-1/2}^{(n+1)^2+1/2} sqrt(x) dx.
>>
>> Now since the derivative of sqrt(x) tends to zero we can
>> use trivial approximations for the parts of the integral
>> at the ends (approximating int_a^b f by (b-a) f(x) for
>> some x in [a,b]) and we get
>>
>> sum_{n^2}^{(n+1)^2} sqrt(i)
>>
>> = o(1) + (n + (n+1))/2 + int_{n^2}^{(n+1)^2} sqrt(x) dx
>>
>> = o(1) + n + 1/2 + (2/3)((n+1)^3 - n^3)
>>
>> = o(1) + n + 1/2 + (2/3)(3n^2 + 3n +1)
>>
>> = o(1) + integer + 1/2 + 2/3
>>
>> = o(1) + integer + 1/6.
>>
>> So the fractional part of the sum tends to 1/6.
>>
>> >Usefull?
>>
>> I can't imagine why. It does seem very curious, though.
>
>So, if k+1 is prime, we get...
>
>lim_{n->inf} FractionalPart(sum_{i=n^k}^{(n+1)^k} i^(1/k))
> = 1/2 - 1/(k+1)

Hmm, looks like we'd get Frac(1/2 + k/(k+1)((n+1)^(k+1)-n^(k+1))),
and if it's true that the binomial coefficient C(k+1,j) is
divisible by k+1 except for j = 0 and j = k+1, which I
imagine is obvious to everyone but me, then we have
Frac(1/2 + k/(k+1)) = 1/2 = 1/(k+1), yes.

Now why is... oops, it's not so. C(4,2) is not divisible
by 4, for example. So for k = 3 we get
Frac(1/2 + 3/4((n+1)^4-n^4)) = Frac(1/2 + (9/2)n^2 +3/4),
which doesn't _have_ a limit as n -> infinity.

Oh wait, 4 is not prime. Yes, if k+1 is prime we
get what you said.

************************

David C. Ullrich

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