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My Problem: Show that the sequence ${x_n}_{n\geq 1}$, where $x_n=(1,\frac{1}{2},\ldots,\frac{1}{n},0,0,\ldots)$ converges to $x=(1,\frac{1}{2},\frac{1}{3},\ldots,\frac{1}{n},\ldots)$ in $l_2$

My Work: Because we are dealing with the space of square summable sequences, we note that: $$d(x,y)=(\sum_{i=1}^\infty |x_i-y_i|^2)^{\frac{1}{2}}$$ Let $\epsilon>0$ be defined. Now we know that by our distance function, $$d(x_n,x)=(\sum_{i=1}^\infty |x_n^i-x_i|^2)^{\frac{1}{2}}=(\sum_{i=n+1}^\infty |x_{n+1}^i-x_i|^2)^{\frac{1}{2}}=(\sum_{i=n+1}^\infty (\frac{1}{i})^2)^{\frac{1}{2}}$$

Now here we use our hint, mainly that $d(x_n,x)\to 0$ as $n\to \infty$. Thus we know that $d(x_n,x)<\epsilon$ for sufficiently large n. Thus $x_n\to x$.

Note: The book gave us that last hint.

My question: I am wondering if this seems like an acceptable answer to this problem. I feel like it is however I am a little skeptical.

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If you assume $d(x_n,x)\rightarrow0$, then you're assuming the result (this is the definition of convergence). You have to prove that the sum on the right hand side of your second displayed equation is eventually bounded by $\epsilon$. But this is easy if you know that $\sum{1\over i^2}$ is convergent. –  David Mitra Jan 21 '13 at 19:37
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The series $\sum_{k=1}^\infty\frac{1}{k^2}$ is convergent, therefore $$ \lim_{n\to \infty}d^2(x,x_n)=\lim_{n \to \infty}\sum_{k=n+1}^\infty\frac{1}{k^2}=0. $$

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