Sum of series and its cubes
Ron Seneketh
A question which a friend of mine posed recently that I've been playing
around with.
Find a sequence a_n (read "a sub n", the nth term of the 'a' sequence) such
that Sum{n=1..infinity}[(a_n)] converges and Sum{n=1..infinity}[(a_n)^3]
diverges (e.g. (a_1)^3 + (a_2)^3 + (a_3)^3 + ... diverges).
The converse is quite simple (i.e., the line (power of one) diverges and the
cube converges). A specific case is a_n = 1/n (since the line will be the
harmonic series and the cube would be a p-series, p = 3 > 1).
Now, some work I have done with this thus far:
Since sum(a_n) converges, lim{n->infinity}a_n = 0 (nth term test).
Therefore, for some N, |a_n| < 1 for n > N. So, since I'm trying to find
a_n and get to define a_n, I may as well say that |a_n| < 1 for all n. Now,
with this statement, this implies that |(a_n)^3| < |a_n| < 1.
Now, if a_n > 0 for all n => (a_n)^3 > 0 for all n. Then, since sum[(a_n)]
converges and |(a_n)^3| = (a_n)^3 < (a_n), this would imply that
sum[(a_n)^3] converged as well (comparison test).
So, a_n must be negative for some values of n. I have tried some simple
alternating series and come up with nothing thus far.
Does anyone have any insights into this problem?
Ron
P.S. - For those of you with Maple, I'm saying that Sum(a[n],n=1..infinity)
converges and Sum((a[n])^3,n=1..infinity) diverges.
Kristal
<ken...@home.com> wrote:
> Hi all,
>
> A question which a friend of mine posed recently that I've been playing
> around with.
>
> Find a sequence a_n (read "a sub n", the nth term of the 'a' sequence) such
> that Sum{n=1..infinity}[(a_n)] converges and Sum{n=1..infinity}[(a_n)^3]
> diverges (e.g. (a_1)^3 + (a_2)^3 + (a_3)^3 + ... diverges).
>
> The converse is quite simple (i.e., the line (power of one) diverges and the
> cube converges). A specific case is a_n = 1/n (since the line will be the
> harmonic series and the cube would be a p-series, p = 3 > 1).
>
For real values define a_n as follows
if i is of the form 3j + 1 a_i = 1/log(j + 2)
if i is of the form 3j + 2 a_i = -1/2(1/log(j + 2)
if i is of the form 3j a_i = -1/2(1/log(j + 3)
The sum of the a_n converges since the terms cancel out in groups of
three and if N is greater than 3m the sum of a_1 +a_2 +... a_N is has
absolute value less than 3/log(m).
The sum of the cubes of the series becomes arbitrarily large as the
groups of three terms no longer cancel out the series becomes
3/4((1/log(2)**3 + 1/log(3)**3 +... which diverges.
Kristal
Virgil
Kri...@yifan.net (Kristal) wrote:
> > Find a sequence a_n (read "a sub n", the nth term of the 'a' sequence) such
> > that Sum{n=1..infinity}[(a_n)] converges and Sum{n=1..infinity}[(a_n)^3]
> > diverges (e.g. (a_1)^3 + (a_2)^3 + (a_3)^3 + ... diverges).
Let w be one of the complex cube roots of 1, so that w^2 + w + 1 = 0.
Let b_n be a sequence for which b_n^3 diverges but b_n -> 0 as n -> +oo.
(for example, b_n = 1/(n+1)^(1/3) ).
Define a_n, zero origin indexed, by:
a_(3*n) = b_n
a_(3*n+1) = b_n*w
a_(3*n+2) = b_n*w^2
Then Sum{i=0..3*n}[a_i] = 0
and Sum{i=1..3*n+1} [a_i]= b_n*w -> 0 as n -> +oo
and Sum{i=1..3*n+2) [a_i]= b_n*(w+w^2) = -b_n -> 0 as n -> +oo
Thus Sum{i=1..+oo)[a_i] = 0.
But w^3 = 1 so (a_3*n)^3 = (a_3*n+1)^3 = a_3*n+2)^3 = b_n,
so obviously Sum{i=1..+oo)[(a_i)^3] diverges with (b_n)^3.
For a real sequence, consider the real part of a_n.
Macavity
"Ron Seneketh" <ken...@home.com> wrote in message
news:jku_6.23353$l45.2...@news20.bellglobal.com...
Hint:
Take the divergent sequence be 1/n, a_n ^3 = 1/n. What all could a_n be?
For ease, you could also write a_n^3 as (-1)^(2n) / n. Do all of the
possible choices for a_n diverge?
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Macavity
"Macavity" <mxy...@hotmail.ignore.com> wrote in message
news:3b3adc45$1...@goliath.newsgroups.com...
Saw Virgil's post, much better. Reminds me of
Ob Puzzle..
Find a closed form for
Sum { over all valid m; (-1)^m C(n, 3m) }
Kristal
<vmh...@home.com> wrote:
> In article <Kristal-3001...@sfr-tgn-sfs-vty68.as.wcom.net>,
> Kri...@yifan.net (Kristal) wrote:
>
> > > Find a sequence a_n (read "a sub n", the nth term of the 'a'
sequence) such
> > > that Sum{n=1..infinity}[(a_n)] converges and Sum{n=1..infinity}[(a_n)^3]
> > > diverges (e.g. (a_1)^3 + (a_2)^3 + (a_3)^3 + ... diverges).
>
> For a real sequence, consider the real part of a_n.
The real parts of 1, w, w^2 are 1, -1/2, -1/2. You have derived a
generalization of my answer to the original problem by in a indirect
manner. It is easier to observe that 1-1/2-1/2 = 0 then that the real
parts of 1 + w + w^2 = 0.
Kristal
Virgil
Kri...@yifan.net (Kristal) wrote:
But my method generalizes nicely to having
(a_1)^n + (a_2)^n + (a_3)^n + ... diverge, for n > 1, but having
(a_1) + (a_2) + (a_3) + ... converge.
Mark Van De Wielle
(that is 1 over the cubic root of n)
define a_n as
a_3n = b_n
a_(3n+1) = -(b_n/2)
a_(3n+2) = -(b_n/2)
sum(a_n) = sum(b_n) + 2*sum((-b_n)/2) = 0
sum((a_n)^3) = sum((b_n)^3) + 2*sum(((-b_n)/2^3))
= sum((b_n)^3) - 1/4*sum((b_n)^3)
= 3/4 sum((b_n)^3)
= 3/4 sum(1/n)
=3/4( 1/2 + 1/3 + 1/4 + 1/5 + ... )
which is known to diverge.
hope it's clear enough, writing down math in ascii symbols isn't all that
simple ;-)
mark
"Ron Seneketh" <ken...@home.com> schreef in bericht
news:jku_6.23353$l45.2...@news20.bellglobal.com...
laurence draper
> say b_n = 1/(n^(-3))
> (that is 1 over the cubic root of n)
i like your solution, but is it possible to make an absolutly convergent
series do this? your's is clearly not.....
Estraven
...
> i like your solution, but is it possible to make an absolutly convergent
> series do this? your's is clearly not.....
No. Seneketh's original post showed that.