Re: divergent series ?

Am 11.12.2006 00:19 schrieb rancid moth:
hello

In W.W. IV, chapter II they have a series of the form

S=1/z + 1/(1+z) +...+ 1/(z+p-1) - 1/(z+p) - 1/(z+p+1) -...- 1/(z+2p+q-1) +
1/(z+2p+q)+...

that is every p-postive set of terms are followed by p+q negative
terms...and it repeats in that fashion. The aim is to show that this series
is divergent. I have an intuitive understanding of why - but im not sure
what would constitute a proof

Consider the finite block os sums of the first p-postivie and p+q negative
terms. re-grouping them one can show that they are

S[1,p,q] = p * Sum(n=0,p-1) 1/((z+n)*(z+p+n) - Sum(n=0,q-1) 1/(z+2p+n)

following on, the second,third... block of sum terms may be written as

S[2,p,q] = p * Sum(n=0,p-1) 1/((z+2p+q+n)*(z+3p+q+n) - Sum(n=0,q-1)
1/(z+4p+q+n)

S[3,p,q] = p * Sum(n=0,p-1) 1/((z+4p+2q+n)*(z+5p+2q+n) - Sum(n=0,q-1)
1/(z+6p+2q+n)

And we see that each finite sum of terms is creating a positive block that
goes like 1/n^2 and a negative block that goes like 1/n.

If we were to add all these blocks then i think this could show that
ultimately we have a series of convergent positive terms, but divergent
negative terms. And so the main series, i think, becomes dominated by the
negative terms which are essentially divergent.

However, even though the summation blocks above are finite the grouping of
the terms in the blocks is a _rearrangement_ in order of the terms in S and
so there is no guarantee that the summation of each finite block will yeild
the same S.

Or,

perhaps it is enough to state that

S[m,p,q] = S[1,p,q]+S[2,p,q]+....+S[m,p,q] ~ -1/m as m becomes large and
therefore so does S ~ O(1/m) and so by the limit comparison theorem is
divergent ?
Hmm, just a guess, don't have time to go deeper in this

if q=0, then if z=1,p=1 we have ln(2)
If we increase z and/or p this should be a multiple/fraction of that,
which would be again convergent, say s_pz.
If q>0 we get for this alternating part another fraction of s_pz, say s_pzq.
But the new negative terms form a fraction of zeta(1), so this partial sum
say s_q, should then be divergent.

But again: just a guess...

Gottfried Helms
.

Relevant Pages

  • divergent series ?
    ... that is every p-postive set of terms are followed by p+q negative ... Consider the finite block os sums of the first p-postivie and p+q negative ... block of sum terms may be written as ... ultimately we have a series of convergent positive terms, but divergent ...
    (sci.math)
  • Re: Series Summing to Different Values
    ... But we can make it sum to ... Note that in each of the two "pools" the terms have infinite sums. ... that such a condition is reached because the negative terms have an ... exceeding -2 at the end of a positive-term run converges downward; ...
    (sci.math)