if the serie of $ u_n $ converges, does it imply that the serie of $ (u_n)^3 $ converges?

Question

if we take a sequence of real numbers ( maybe we can also think on this question if the numbers are complex ? ), does $$\sum_{n=0}^\infty u_n $$ converges, imply that $$\sum_{n=0}^\infty (u_n)^3 $$ converges also

happy new year btw !

Answer 1

Here's a standard counterexample . . .

For each positive integer $n$, let $x_n = {\large{\frac{1}{\sqrt[3]{n}}}}$.

Consider the series $$ \left(x_1 - {\small{\frac{1}{2}}}x_1 - {\small{\frac{1}{2}}}x_1\right) +\left(x_2 - {\small{\frac{1}{2}}}x_2 - {\small{\frac{1}{2}}}x_2\right) +\left(x_3 - {\small{\frac{1}{2}}}x_3 - {\small{\frac{1}{2}}}x_3\right) + \cdots $$ with the parentheses removed (they're shown only to make the pattern clear).

All partial sums are nonnegative, and the sum of the first $m$ terms is less than ${\large{\frac{1}{\sqrt[3]{n}}}}$ if $m > 3n-2$.

It follows that the series sums to zero.

But the series $$ \left(x_1^3 - {\small{\frac{1}{8}}}x_1^3 - {\small{\frac{1}{8}}}x_1^3\right) +\left(x_2^3 - {\small{\frac{1}{8}}}x_2^3 - {\small{\frac{1}{8}}}x_2^3\right) +\left(x_3^3 - {\small{\frac{1}{8}}}x_3^3 - {\small{\frac{1}{8}}}x_3^3\right) +\cdots $$ diverges, since the sum of the first $3n$ terms is $$ {\small{\frac{3}{4}}} \left( 1+{\small{\frac{1}{2}}}+{\small{\frac{1}{3}}} +\cdots +{\small{\frac{1}{n}}} \right) $$

Answer 2

Set $S_{m,l}=\sum_{k=m}^l\frac{1}{k}$. Find $m,l$ for $S_{m,l}=\sum_{k=1}^l\frac{1}{k}$ such that $n^{-\frac{1}{3}}\leq S_{m,l}\leq n^{-\frac{1}{3}}+2^{-n} $.

We define the $a_i=\frac{1}{k-1+i}$ for $1\leq i\leq l-k+1$, $a_{l-k+2}=-n^{-\frac{1}{3}}.$ We repeat this procedure to define $a_i$ for all $i$.

Then $\sum a_i$ converges. Then define $p_i,r_i$ the positive and negative terms respectively.

We have that $\sum_{i}p_i^3<\sum_{i}\frac{1}{n^3}$, so the positive terms converge.

However, $\sum_{i}r_i^3=-\sum_{i}\frac{1}{i}$, so the negative terms diverge.

Therefore, $\sum_{i}a_i^3$ is not convergent.

Idea: First, we observe that $u_n$ cannot keep a sign. So the sum of $u_n$ must have negative and positive mass, each adding to infinity, canceling each other out.

Now, the idea is to use the convexity of $f(x)=x^3$ when restricted on the positive reals. More specifically $f$ is supper additive i.e. $f(\sum_i x_i)\geq \sum_if( x_i)$. So, spreading the mass and applying $f$ gives something small, compared to the same mass if it was concentrated.

So, we want to spread the positive mass, while having the negative mass concentrated. By doing this appropriately we should be able to construct a series $u_n$ such that the $u_n^3$ series has the positive sum converging and the negative sum diverging.