As title says does the converges of $\sum_{i=0}^\infty a_i$ imply that $\sum_{i=0}^\infty (a_i)^3$ converge?
Asked
Active
Viewed 789 times
3 Answers
6
If we assume $a_i\leq 0$ or $a_i\geq 0$ from some point on the statement is obviously true.
If we remove the crucial assumption about the sign, it might not. Let
$$ a_n = \frac{1}{\log^2(n+2)}\cdot\left\{\begin{array}{lcl}1&\text{if}&n\equiv 1\pmod{3}\\1&\text{if}&n\equiv 2\pmod{3}\\-2&\text{if}&n\equiv 0\pmod{3}\end{array}\right.$$
It it simple to check that $\sum_{n\geq 0}a_n$ is convergent, but $\sum_{n\geq 0}a_n^3$ is not, essentially by Kronecker's lemma (summation by parts).
Jack D'Aurizio
- 361,689
-
1Nice. I tried $a_n = \frac{1}{n^{1/3}}$ with the same $+1,+1,-2$ and this should also work. – Winther Jun 19 '19 at 19:37
-
@Winther: agreed. It is enough to have $b_n>0$ such that $\sum b_n^3$ is divergent but $\sum b_n/n$ is convergent. – Jack D'Aurizio Jun 19 '19 at 19:38
-
Thank you very much! I never even heard of this lemma nor did i think about this kind of counter-point. – Dustbony Jun 19 '19 at 19:48
4
Let's define the sequence by $a_{3k-2}=2/\sqrt[3]{k}$, $a_{3k-1}=a_{3k}=-1/\sqrt[3]{k}$, note that the sum of these three consecutive terms is $0$. But $a_{3k-2}^3+a_{3k-1}^3+a_{3k}^3=6/k$
From here, it's easy to check that $\sum a_n=0$ but $\sum a_n^3$ diverges.
Julian Mejia
- 4,458
-
1You may want to start with $n=4$ because if $n=1,2,3$ then $k=0$. Otherwise, change the indexing of $a_n$. – Somos Jun 19 '19 at 20:13
-1
$\textbf{Hint: }$ If $$\sum_{n=1}^{\infty}a_n$$ converge, then $a_n\to 0$ as $n\to \infty$. Which means, there exists some $N\in \mathbb{N}$ such that $|a_n|<1$ for all $n\ge N$. Then $|a_n|^3=|a_n^3|<|a_n|$, and hence $$\sum_{n=N}^\infty a_n^3$$ converge (the absolute convergence is necessary).
azif00
- 23,123