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?


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.


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 $\tilde S^*$.


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.


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