Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove or disprove:

Let $\rho : \mathbb{N} \rightarrow \mathbb{N}$ injective. Let $(a_{n})_{n \in \mathbb{N}}$ be a sequence.

(i) If $\displaystyle\sum\limits_{n=1}^\infty a_{n}$ absolutely converges then $\displaystyle\sum\limits_{n=1}^\infty a_{\rho(n)}$ also converges absolutely.

(ii) If $\displaystyle\sum\limits_{n=1}^\infty a_{n}$ converges then $\displaystyle\sum\limits_{n=1}^\infty a_{\rho(n)}$ also converges.

So my understanding is that a series converges if the infinite sum of the series is the limit? It converges absolutely if $\displaystyle\sum\limits_{n=1}^\infty |a_{n}|$ converges. The way I am reading the $a_{\rho(n)}$ 's is that they denote some sort of permutation or rearrangement of the original series. Aside from that though, I am absolutely lost when it comes to approaching this problem... In general am I looking for an $\epsilon >0$ which is greater than $a_{\rho(1)}+ \dots +a_{\rho(n)}$ ? If so, how do I begin trying to find it?

share|improve this question
    
You may also be interested in the Riemann series theorem, mathworld.wolfram.com/RiemannSeriesTheorem.html and its extensions due to Sierpinski, that came up recently in a question on MO: mathoverflow.net/questions/47589/… –  Andres Caicedo Dec 3 '10 at 3:37

3 Answers 3

up vote 9 down vote accepted

For (i):

Suppose that $\sum a_n$ converges to $a \in \mathbb{R}$. So if $\epsilon > 0$ there exists N such that if $n, l > N$ and $s_n = a_1 + \ldots + a_n$ then

$$|a - s_n| < \epsilon \text{ and } \sum_{k = N + 1}^l |a_n| < \epsilon.$$

Now let $M \in \mathbb N$ such that the terms $a_1, \ldots, a_N$ are contained as sum elements in $t_m = a_{\rho(1)} + \ldots + a_{\rho(M)}$. So now we have that for $m \geq M$ that $t_m - s_n$ is a sum of finitely many terms $a_l$ with $l > N$. So, for some $l > N$ we have that

$$|t_m - s_n| \leq \sum_{k = N + 1}^l |a_n| < \epsilon.$$

So for $m \geq M$ we have:

$$|t_m - a| \leq |t_m - s_n| + |s_n - a| < \epsilon + \epsilon = 2\epsilon.$$

So the rearrangement converges.

(ii) is not true. Let $\sum_n (-1)^n/n$ then take the $\rho$ which maps the integers injectively to the even integers.

share|improve this answer

I just want to give an alternative proof for (i).

To say that $\rho : \mathbb{N} \rightarrow \mathbb{N}$ is injective means that for every $n \in \mathbb{N}$, $\rho(n) \in \mathbb{N}$, and if $n \neq m$, then $\rho(n) \neq \rho(m)$. Define $S^*_n = |a_1| + |a_2| + \cdots + |a_n|$, and $\tilde S^*_n = |a_{\rho(1)}| + |a_{\rho(2)}| + \cdots + |a_{\rho(n)}|$. Obviously, since $\rho$ is injective, $\tilde S^*_n \leq |a_1| + |a_2| + |a_3| + \cdots$. But the right-hand side exists as a finite nonnegative number (since $\displaystyle\sum\nolimits_{n=1}^\infty a_{n}$ absolutely converges, or, equivalently, $S^*_n$ converges). So, $\tilde S^*_n$ is a monotone increasing sequence, bounded from above by $\sum\nolimits_{n = 1}^\infty {|a_n |}$. Hence, $\tilde S^*_n$ converges to a finite nonnegative number $\tilde S^* \leq \sum\nolimits_{n = 1}^\infty {|a_n |}$. That is, $\displaystyle\sum\nolimits_{n=1}^\infty a_{\rho(n)}$ converges absolutely to $\tilde S^*$.

share|improve this answer

Here is a third alternative to statement (i).

Let $A=\sum_{n=1}^\infty |a_n|$ and consider the sequence of partial sums $S_N=\sum_{n=1}^N |a_{\rho(n)}|$.

(1) Since $\rho$ is injective $S_N\le A$ for all $N$.

(2) Note that $S_N\le S_{N+1}$ for all $N$.

By (1) $S_N$ has a finite least upper bound (that is $\sup S_N<\infty$), and by (2) the sequence grows to that bound.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.