Is the Series Sum of Cubes Necessarily Convergent?

[lukepeterpaul]

Hi,
"Given that sum x_n converges, where x_n are real, does sum (x_n)^3 necessarily converge?"
My gut feeling is "no". When considering conditional convergent series. It may be that the cubing can increase the ratio of consecutive "groups of terms" (ie the terms in the series we consider as one during the ratio test) s.t. the series is no longer convergent.
M I right?
If so, how might I go about constructing an example?
Thanks!

 

[AlephZero]

The sum of x_n converges,

Therefore, there is a value of N such that |x_n| < 1/2 for all n > N

For all n > N, (x_n)^3 is closer to 0 than x_n/4

Therefore the sum of (x_n)^3 converges.

 

[g_edgar]

Aleph:
That argument is correct for nonnegative series, but not for general convergent series.

 

[lukepeterpaul]

I agree with g_edgar, hence I mentioned "conditional" convergence. :)

 

[yonat83]

for all n define the finite sequence of n+1 elements

1/n^(1/3) , 1/(n*n^(1/3)), ..., 1/(n*n^(1/3))

when ... means n times 1/(n*n^(1/3)).

for all n, put those sequences on a row and look at the sequence a_i you get.

then sum (a_i) converges since for all n sum(a_i) < 1/(n*n^(1/3)) after some index N.
But sum(a_i^3) doesn't converge since the sum of cubes of any "short" sequence of index n is

1/n - 1/n^4 - ... - 1/n^4 = 1/n - 1/n^3 which behaves like 1/n as n grows

Thus sum(a_i) behaves like the harmonic series sum(1/n) which diverges.

 

[yonat83]

Sorry, meant 1/n^(1/3) , -1/(n*n^(1/3)), ..., -1/(n*n^(1/3))

with n times -1/(n*n^(1/3))

 

[AlephZero]

Well, I guess my mistake had some educational value, since the counter example answers the OP's question