Re: Accidental limit

From: The Last Danish Pastry (TheLastDanishPastry_at_yahoo.com)
Date: 06/02/04 

Date: Wed, 2 Jun 2004 15:36:01 +0100

"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) 

-- 
Clive Tooth
http://www.clivetooth.dk

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