Re: 1+2+3+.....
- From: Narcoleptic Insomniac <i_have_narcoleptic_insomnia@xxxxxxxxx>
- Date: Thu, 26 Jan 2006 14:34:55 EST
On Jan 26, 2006 12:16 PM CT, gwlucky@xxxxxxxxxx wrote:
> Here's an easier example to show the danger of
> dealing with non-convergent series and moving
> parentheses:
>
> Consider the series
>
> 1 -1 +1 -1 +1...
>
> If we write
>
> (1 -1) + (1 -1)...
>
> it "appears" that the sum is zero.
>
> But if we write
>
> 1 + (-1 +1) + (-1 +1)...
>
> is "appears" that the sum is 1.
>
> You must have convergence and absolute convergence to
> be able to move
> terms around and add in any order you want.
>
> I think it was Jakob Bernoulli who established in a
> remarkable proof even to this this day that
>
> 1 + 1/2^2 + 1/3^2 + 1/4^2 ... = pi^2/6
I'm pretty sure that it was Euler in 1739 and that he
did it by considering the infinite product representation
of the sine function.
> A description of his proof may be found in "A Journey
> Through Genius" by Dunham (Wiley Press).
>
> Now, Jakob broke the rules since the proof relies on
> manipulating two infinite series know to equal each
> other. This was before the time of the formalizam of
> Weierstrass (epsilon/delta). It turns out he was
> "absolutely" justified in what he did, however.
>
> Bernoulli actually went on up through
>
> 1 + 1/2^26 + 1/3^26... = pi^26/integer
Again, I believe that this was in fact Euler. However,
the coefficients of these (even zeta function) sums can
be given in terms of Bernoulli numbers.
> but he never establish the result for odd powers,
> such as
>
> 1 + 1/2^3 + 1/3^3...
>
> and in fact it has never been established that the
> sum may be written as
>
> pi^3/integer
Correct, in fact, I've been somewhat obsessed lately with
obtaining a closed form for Aprey's contstant *_*
On a topic more related to the thread, the Riemann series
theorem says that if a series
\sum_{n = 0}^{\infty} a_n
...converges, but the series...
\sum_{n = 0}^{\infty} |a_n|
...diverges, then the series is conditionally convergent
and the terms may be rearranged to convege to any value.
Regards,
Kyle
.
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