Show that $(\mathbb{R}^\infty,\lVert\rVert)$ is not complete space. 
Let $\mathbb{R}^\infty$ the vector space of sequences $(x_n)_{n\in\mathbb{N}}$ of real numbers, such that has a finitely many terms $x_n\neq 0$. Define on $\mathbb{R}^{\infty}$ the norm $\lVert x\rVert=\sqrt{\sum_{i=1}^\infty x_i^2}$, for $x=(x_n)_{n\in\mathbb{N}}$. Show that $(\mathbb{R}^\infty,\lVert\cdot\rVert)$ is not complete space. 

The prove of this question -is similar to show that $\ell^{2}$ with the norm $\lVert x\rVert=\sqrt{\sum_{i=1}^\infty x_i^2}$, for $x=(x_n)_{n\in\mathbb{N}}$, is a complete metric space-. But, how the fact that "has a finitely many terms $\neq 0$" implies that space isn't complete?  
 A: Consider first the sequence $u=(u_n)_{n\in\mathbb{N}}$ defined by $u_n = \frac{1}{n+1}$ (which is not in $\mathbb{R}^\infty$). From it, define the sequences
$
u^{(k)} = (u^{(k)}_n)_{n\in\mathbb{N}}
$
(for $k\in\mathbb{N}$) as the truncations of $u$ to the first terms:
$$
u^{(k)}_n = \begin{cases}
u_n & \text{ if } n \leq k \\
0 & \text{ otherwise.} \\
\end{cases}
$$
Then, you have $$\sum_{n=1}^\infty (u_n - u^{(k)}_n)^2 = \sum_{n=k+1}^\infty \frac{1}{(n+1)^2} \xrightarrow[k\to\infty]{} 0$$ so $(u^{(k)})_k$ converges to $u$ in $\ell^2$, and therefore is Cauchy. But then, it is also Cauchy in $(\mathbb{R}^\infty, \lVert\cdot\rVert)$ which is a subspace of $\ell^2$; yet it cannot converge in $(\mathbb{R}^\infty, \lVert\cdot\rVert)$, since $u\notin \mathbb{R}^\infty$.
A: Let $e_m^*(e_k) =  \delta_{mk}$, that is, the linear functional that picks
out the $m$th component. It is easy to check that $e_m^*$ is continuous.
Suppose $x_n \to x$ in $\mathbb{R}^\infty$. Since $x \in \mathbb{R}^\infty$,
there is some $N$ such that $e_N^*(x) = 0$, and hence $e_N^*(x_n) \to 0$.
Now let $x_n = \sum_{k=1}^n{1 \over 2^k} e_k$. It is easy to check that $x_n$
is Cauchy.
Choose some $m$, then we see that $e_m^*(x_n) = {1 \over 2^m}$
for all $n \ge m$.
Hence $x_n$ cannot converge to a point in $\mathbb{R}^\infty$.
A: $$
\left( 1,  \frac 1 2 , \frac 1 3, \ldots,\frac 1 n , 0 , 0, 0, 0, \ldots \right) \to\text{what} \in \ell^2 \text{ ?}
$$
The point is you can have a sequence within the space of sequences with only finitely many nonzero terms, whose limit in $\ell^2$ requires infinitely many nonzero terms, and is thus not within that smaller space.  The smaller space is therefore not complete.
