Proving or refuting declarations connected to convergent and also definitely convergent collection

Prove or refute :

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

(i) If $\displaystyle\sum\limits_{n=1}^\infty a_{n}$ definitely merges after that $\displaystyle\sum\limits_{n=1}^\infty a_{\rho(n)}$ additionally merges definitely.

(ii) If $\displaystyle\sum\limits_{n=1}^\infty a_{n}$ merges after that $\displaystyle\sum\limits_{n=1}^\infty a_{\rho(n)}$ additionally merges.

So my understanding is that a collection merges if the boundless amount of the collection is the restriction? It merges definitely if $\displaystyle\sum\limits_{n=1}^\infty |a_{n}|$ merges. The means I read the $a_{\rho(n)}$'s is that they represent some type of permutation or reformation of the initial collection. In addition to that though, I am definitely shed when it involves approaching this trouble ... In basic am I seeking an $\epsilon >0$ which is more than $a_{\rho(1)}+ \dots +a_{\rho(n)}$? If so, just how do I begin searching for it?

2019-12-02 02:52:59
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Answers: 3

Here is a 3rd choice to declaration (i).

Allow $A=\sum_{n=1}^\infty |a_n|$ and also take into consideration the series of partial amounts $S_N=\sum_{n=1}^N |a_{\rho(n)}|$.

(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 limited the very least upper bound (that is $\sup S_N<\infty$), and also by (2) the series expands to that bound.

2019-12-03 05:39:19

For (i) :

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

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

Now allow $M \in \mathbb N$ such that the terms $a_1, \ldots, a_N$ are had as amount components in $t_m = a_{\rho(1)} + \ldots + a_{\rho(M)}$. So currently we have that for $m \geq M$ that $t_m - s_n$ is an amount of finitely several 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 reformation merges.

(ii) is not real. Allow $\sum_n (-1)^n/n$ after that take the $\rho$ which maps the integers injectively to the also integers.

2019-12-03 04:22:13

I simply intend to offer a different evidence for (i).

To claim that $\rho : \mathbb{N} \rightarrow \mathbb{N}$ is injective methods that for every single $n \in \mathbb{N}$, $\rho(n) \in \mathbb{N}$, and also if $n \neq m$, after that $\rho(n) \neq \rho(m)$. Specify $S^*_n = |a_1| + |a_2| + \cdots + |a_n|$, and also $\tilde S^*_n = |a_{\rho(1)}| + |a_{\rho(2)}| + \cdots + |a_{\rho(n)}|$. Clearly, given that $\rho$ is injective, $\tilde S^*_n \leq |a_1| + |a_2| + |a_3| + \cdots$. Yet the appropriate - hand side exists as a limited nonnegative number (given that $\displaystyle\sum\nolimits_{n=1}^\infty a_{n}$ definitely merges, or, equivalently, $S^*_n$ merges). So, $\tilde S^*_n$ is a monotone raising series, bounded from above by $\sum\nolimits_{n = 1}^\infty {|a_n |}$. Therefore, $\tilde S^*_n$ merges to a limited nonnegative number $\tilde S^* \leq \sum\nolimits_{n = 1}^\infty {|a_n |}$. That is, $\displaystyle\sum\nolimits_{n=1}^\infty a_{\rho(n)}$ merges definitely to $\tilde S^*$.

2019-12-03 04:21:58