0
$\begingroup$

I want to prove convergence of sequence.

Let $a_k$ be a sequence in $\mathbb{R}^k$ with $a_k$=$(a_{ki}...a_{kn})$ for all k element of natural numbers, and let $a=(a_1,...a_n)$ element of $\mathbb{R^n}$

Then ($\overset{\rightharpoonup }{a_k} \to \overset{\rightharpoonup } a$) iff for each i $a_{ki}$ approches $a_{i}$

Equivalent to the convergence of each of its coordinate sequences.

To prove this:

$\left|\left|a_k-a\right|\right|{}^{\wedge}2$ = $\sum _{i=1}^n \left(a_{\text{ki}}-a_i\right){}^2$

Suppose $a_{ki} ->a_i$ for each i=1,...,n, Choose any $\epsilon $,

to each i=1,...,n there corresponds a number $k_i$

$\left|a_{\text{ki}}-a_i\right|{}$ $<\sqrt{\frac{\epsilon }{n}}$ whenever $k>K_i$ element of naturals

Let $K=max(K_i)$

then $\left|\left|a_k-a\right|\right|{}^{\wedge}2<\epsilon$ Therefor converges to zero, hence $\left|a_{\text{ki}}-a_i\right|{}$ converges to zero.

Now since this is iff, how do I prove the converse?

$\endgroup$
5
  • $\begingroup$ The sentence "For all $k>=K\in\mathbb{N}$ then ($\overset{\rightharpoonup }{a_k} \to \overset{\rightharpoonup } a$) iff for each i $a_{ki}$ approches $a_{i}$ " is gibberish. Can you please write down, clearly, (1) what you know and (2) what you want to prove? $\endgroup$
    – 5xum
    Oct 9, 2018 at 11:23
  • $\begingroup$ Is this better? @5xum $\endgroup$
    – ALEXANDER
    Oct 9, 2018 at 11:34
  • $\begingroup$ Much. Unlike before, people (not just you) can now actually understand what you are asking. $\endgroup$
    – 5xum
    Oct 9, 2018 at 11:37
  • $\begingroup$ However, I still don't understand your proof. I don't even know the direction you are proving. In your proof, if you want anyone to understand it, you must clearly write what you are proving, and which statement follows from which statement. Your "proof" is just a mess of equations and inequalities, and I don't even know what follows from what. I don't know what $K_i$ is, or where it came from, I don't know what your assumption is, what your conclusion is, I don't know what $\epsilon$ is, or where it came from... $\endgroup$
    – 5xum
    Oct 9, 2018 at 11:38
  • $\begingroup$ @5xum Does the update clarify the direction? $\endgroup$
    – ALEXANDER
    Oct 9, 2018 at 11:46

2 Answers 2

3
$\begingroup$

I give here both if part and only if part.

Only If Part: Let $\left\{\overrightarrow {a_k}\right\}_k$ converges to $\overrightarrow a$ in $\Bbb R^n$. Then using norm $||\cdot||$ of $\Bbb R^n$ we can say that given any $\varepsilon > 0$ we have a natural number $k_0$ such that whenever $k≥k_0$ we have $ \left|\left|\overrightarrow {a_k} -\overrightarrow a\right|\right| < \varepsilon$ i.e. $\displaystyle\sum _{i=1}^n \left(a_{\text{ki}}-a_i\right){}^2 <\varepsilon^2$ for $k≥k_0$. Since each term of summation is non-negative we have $|a_{ki} - a_i|<\epsilon$ for each $i=1,2,...,n$ and for $k≥k_0$. Therefore the $a_{ki}\rightarrow a_i$ for each $i=1,2,...,n$.

If Part: Conversely assume for a sequence $\left\{\overrightarrow {a_k}\right\}_k$ and a vector $\overrightarrow a$ in $\Bbb R^n$ we have $a_{ki}\rightarrow a_i$ for each $i=1,2,...,n$. Then for given any $\varepsilon >0$ each $i=1,2,...,n$ choose natural numbers $m_1,m_2,...,m_n$ such that whenever $k≥m_i$ we have $|a_{ki}-a_i|< {\frac{\varepsilon}{\sqrt n}}$ for each $i=1,2,...,n$. Then choose $k_0=\max\{m_1,m_2,...,m_n\}$ and now for $k≥k_0$ we get $\displaystyle\sum _{i=1}^n \left(a_{\text{ki}}-a_i\right){}^2<\varepsilon ^2$ i.e. for $k≥k_0$ we have $\left|\left|\overrightarrow {a_k} - \overrightarrow a\right|\right|<\varepsilon$ i.e. $\left\{\overrightarrow {a_k}\right\}_k \rightarrow \overrightarrow a$ as $k\rightarrow \infty$.

$\endgroup$
1
$\begingroup$

Hint: To prove the iff side, use the inequality $\left|\left|a_k-a\right|\right|{} \geq \left|a_{\text{ki}}-a_i\right|{}$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .