Prove that is divergent.

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- May 22nd 2009, 09:19 PMNonCommAlgInfinite series (1)
Prove that is divergent.

- May 24th 2009, 06:45 AMDeadstar
Does it diverge? According to maple it converges to 1.412025371...

- May 24th 2009, 11:02 AMNonCommAlg
- May 26th 2009, 09:43 AMchisigma
The original series is...

(1)

... and it can be transformed in the form...

(2)

... where is only changing the order of its terms. But (1) is not absolutely convergent and in this case changing the order of its terms can transform the series from convergent to divergent or vice-versa, so that (Thinking)...

Kind regards

- May 26th 2009, 10:13 AMOpalg
... but if we keep the terms in the same order then we won't alter the convergence/divergence. Roughly speaking, the terms of the series have the same sign throughout any interval of the form . The sum of the absolute values of the terms in that interval is So the series alternates between a batch of positive terms adding up to approximately1, followed by a batch of negative terms adding up to approximately –1. This oscillation means that it cannot converge.

- May 26th 2009, 12:31 PMchisigma
The series...

(1)

... properly speaking is neither convergent nor divergent, it is*indeterminated*[for someone it converges to but is not the case to discuss about that here...]. If we 'alterate' in some way [also in 'microscopic way'...] the (1), it can...

a) 'explode' [diverge]

b) 'implode' [converge]

c) remain 'stable' [indeterminated]

In the case...

(2)

... what of the hypothesis a),b), c) is 'true'?...

Kind regards

- May 27th 2009, 02:01 AMchisigma
Considering the identity...

(1)

... where the symbol means 'fractional part of', the proposed series can be written as...

(2)

Now if we consider the sequence , it generates the series...

(3)

... that is convergent since the function is analytic in the whole complex plane with the only exception of the point . So if the sequence produces a convergent series, may be not hazarded suppose that the sequence produces also a convergent series. It is obvious that this conclusion must be supported in rigorous fashion... at first however seems to be not an absurd 'a priori'...

Kind regards

- May 27th 2009, 12:33 PMOpalg
The zeta function is analytic in , but it is

*not*equal to the sum of the series on the whole of that domain. In fact, the series only converges in the region . To extend it beyond there, you have to use analytic continuation. The equation (3) is not correct because the left side does not converge. - April 6th 2010, 05:54 PMchiph588@
- July 4th 2010, 03:24 PMchiph588@
- July 4th 2010, 10:23 PMCaptainBlack
- July 4th 2010, 10:26 PMCaptainBlack
- July 5th 2010, 10:37 PMchisigma
The following is from 'T. J. A. Bromwich An introduction to the theory of infinite series [London: Macmillan, 1908]' ...

http://digilander.libero.it/luposabatini/Bromwich2.JPG

The [british] term*oscillatory*is much better than the [italic]*indeterminate*to indicate a*bounded*sequence that has no finite limit, so that I'll be glad to use it in the future (Itwasntme)...

Kind regards

- July 6th 2010, 12:07 AMOpalg
- July 6th 2010, 08:29 AMchiph588@