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In $(\ell ^\infty,{\Vert .\Vert_\infty)}$, how would I show that $x_n=\left(\frac{n+1}{n},\frac{n+2}{2n},\frac{n+3}{3n}, ...\right)$ converges and how would I find the limit?

I tried using the fact that the uniform norm ${\Vert .\Vert_\infty}= \text{sup}|X_n|$ and the definition of convergence is that given $\epsilon > 0$, there exists $N \in \mathbb N$ such that $d(x_n,x)< \epsilon$ for all $n>N$, but I cant seem to show it converges. How would I show it converges and find the limit?

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If it converges, then it will converge pointwise, i.e., in every coordinate. This helps to find the limit. –  abatkai Feb 19 '12 at 22:51
    
@abatkai, Hi, thanks for your reply. I haven't really covered pointwise limits etc. I think I was meant to show it converges using the definition of convergence. Is it possible to do it working from the definition? –  Aaron Feb 19 '12 at 23:00
    
Yes, but you need first a candidate for the limit; otherwise you cannot use the definition. –  abatkai Feb 19 '12 at 23:03
    
To use the definition, you need to know what the limit is. @abatkai was suggesting a way to determine that: find the limit of the first coordinates, find the limit of the second coordinates, etc., form a sequence $x$ with those limits as its coordinates, and then use the definition to show that $x_n\to x$. –  Brian M. Scott Feb 19 '12 at 23:03
    
What you get is what chessmath answered. –  abatkai Feb 19 '12 at 23:04

2 Answers 2

up vote 1 down vote accepted

We denote $x^{(n)}$ the sequence of sequences. Then $x_k^{(n)}=\frac{n+k}{kn}=\frac 1k+\frac 1n$, and we can write $x^{(n)}=a^{(n)}+b^{(n)}$ where $a^{(n)}_k=\frac 1k$ for all $n$ and $b_k^{(n)}=\frac 1n$. We have $||b^{(n)}||_{\infty}=\frac 1n$ which converges to $0$ and $a^{(n)}$ doesn't depend on $n$. Denoting this sequence $a$, we can see that $\lim_{n\to\infty}||x^{(n)}-a||_{\infty}=\lim_{n\to\infty}||b^{(n)}||_{\infty}=0$.

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Thanks Davide, interesting to see a different way of writing it. –  Aaron Feb 19 '12 at 23:42

Try the limit $(1,1/2,1/3,..., 1/k,...)$.

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Thanks for your help, most appreciated. –  Aaron Feb 19 '12 at 23:18

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