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.
