Re: divergent series

On Jan 5, 3:07 pm, Denis Feldmann <feldmann.denis.asuppri...@xxxxxxx>
wrote:
It is well known that wildly divergent series (like 1+2+3+4+...) can be
"summed" by various means (Cesaro, Abel, Dirichlet...) which usually
respect some rules (like linearity) and extend the usual convergence (it
is also possible to change the underlying topology, like with p-adic
metrics, but this is not what I am interested in here). On all this,
the following Wikipedia introduction can be useful :http://en.wikipedia.org/wiki/Divergent_series

But what can one think of the following "proof" : let's, for simplicity,
note a0+a1+a2+... = S to say that some summation process P gives S
when applied to the series sum(a_n) (i.e. P((sum(a_n))=S), and assume P
is linear and stable, i.e. a0+a1+a2+... =a0+ (a1+a2+a3...), we get
then (if x<>1) S=1+x+x2+...=1+ (x+x2+...)=1+x(1+x+x2+...)=1+xS =>
S=1/(1-x), so (with x=-1) 1-1+1-1+...=1/2 , then
T=1-2+3-4+...=1-(2-3+4-...)=1-((1-2+3-...)+ (1-1+1-1+...))=1-(T+1/2)=>
T=1/4 and at last U=1+2+3+...=(1-2+3-4+ ...)+ (0+4+0+8+0+12+...)=1/4+4U,
therefore U=-1/12, which indeed is zeta(-1).

Well, everything seems to fit well, except, alas, that P cannot be
stable : one should then get X=1+1+1+...=1+(1+1+1...)=1+X, which is
annoying, to say the least (otoh, 1+1+1+...=1-1+1-1+...+ (0+2+0+2+...)
getting to the correct value 1+1+1+...=zeta(0)=-1/2)

Therefore, I must be doing something right (as the whole trick works as
soon as one get plausible equations) but what is it, exactly?

One problem with the notation you introduce with

let's, for simplicity,
note a0+a1+a2+... = S to say that some summation process P gives S

is that the right hand side is not well defined: there
exist sequences (a_i) and summation processes which
are linear, stable and which coincide with the usual
summation for convergent series, which assign to
sum(a_i) different values.

Therefore your = relation is not transitive...

(One way to obtain summation procedures is the following
(which for simplicity I only do for series with bounded
partial sums:

Let S be the real vector space of one-sided sequences of
real numbers which are bounded, normed with the supremum
norm, and let T : S -> S be the linear map which drops
the 1st item in a sequence and shifts everything else
one space back. Let C be the subset of S of convergent
sequences and let L : C -> R be the linear form which maps
a convergent sequence to its limit. Then of course
L(T(a)) = L a whenever a is a convergent sequence, and
L is a bounded form on the vector space C. Now, using
a variant of the Hahn-Banach theorem, one sees that there
exists extensions L' of L from C to the whole space S, which
still satisfy that L' composed with T coincides with L',
and which is still bounded.

Now e can define a summation procedure for series sum(a_n)
with bounded partial sums: simply let s=(s_n) be the
sequence of partial sums of sum(a_n) and say that the `sum'
of sum(a_n) is L'(s).

One can see without much trouble that the extension L'
is _not_ unique, therefore one obtains in this way
summation procedures which give some series different
values.)

-- m
.

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