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?
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 $k\rightarrow \infty$.