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Given any (fixed) element $x\in l_1$, show that the sequence $x ^{\left( k \right)} = \left( x _{1} , . . . , x _{k} , 0 , . . . \right)\in l_1$ (i.e. the first $k$ terms of $x$ followed by all $0s$) converges to x in $l_1-norm$. Show that the same holds true in $l_2$, and gives an example showing that it fails in $l_ {\infty}$.

$l_1-norm = \sum_{n=1}^{\infty}|x_n|$

$l_p$ is the collection of all real sequences $x=(x_n)$ for which $\sum _{n=1}^{\infty }|x_n|^p < {\infty }$.

$l_{\infty}$ is the collection of all bounded real sequences.

Actually, I did't understand this question. I didn't know what the limit x is. So, I didn't know how to prove this problem.

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    $\begingroup$ $x$ is an element of $l^1$, and $\{x^{(k)}\}$ is a sequence of elements in $l^1$, the question is asking to show that $\|x^{(k)} - x\|_1$ goes to zero as $k$ approaches infinity. $\endgroup$ – Xiao Sep 17 '16 at 21:09
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The question is asking you to prove that the sequence $x^{(k)}$ converges to $x$ in the sense that for every $\epsilon > 0$, $||x-x^{(k)}||_{l_1} < \epsilon$ for all $k$ past a certain point.

Note $||x-x^{(k)}||_{l_1} = \sum_{n=k+1}^{\infty} x_n$. But we are given $x \in l_1$ so that $\sum_{n=1}^{\infty} x_n$ converges. What does this tell you about sums of the form $\sum_{n=k+1}^{\infty} x_n$ for large $k$?

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