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?  

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 $u\notin \mathbb{R}^\infty$.
