Prove that the convergent sum of a real sequence is a metric I want to show that
$$
\varrho(\{a_n\},\{b_n\})=\left(\sum_{n=0}^\infty{(a_n-b_n)^2}\right)^{1/2} 
$$
is a metric, where $\{a_n\}_{n\in\Bbb N}\in \ell_2$, and $\ell_2$ is the set of all real sequences $\{a_n\}_{n\in\Bbb N}$ such that $\sum_{n=0}^\infty{a_n^2}$ converges.

So far, I have the following:
To demonstrate that $\varrho$ defines a metric, it must satisfy four conditions: non-negativity, identity of indiscernibles, symmetry, and the triangle inequality. These conditions will be demonstrated below.
Non-negativity:
It can be clearly seen that $\varrho$ will be non-negative since $(a_n-b_n)^2$ will be positive, $\Sigma$ is the sum of these positive values, and the square root of this sum will also be positive.
Identity of Indiscernibles:
It can be seen that $\varrho(\{a_n\},\{b_n\})=0$ only when $(a_n-b_n)^2=0$, whence $a_n=b_n$ for every $n\in\Bbb N$.
Symmetry:
Due to the form of the function, we can remove a factor of $(-1)$ from within to give
$$
\varrho(\{a_n\},\{b_n\})=\left(\sum_{n=0}^\infty{\left((-1)(b_n-a_n)\right)^2}\right)^{1/2}=\left(\sum_{n=0}^\infty{(b_n-a_n)^2}\right)^{1/2}=\varrho(\{b_n\},\{a_n\})
$$
Thus $\varrho$ also has symmetry.
Triangle Inequality:
We want to show that 
$$
\varrho(\{a_n\},\{c_n\})+\varrho(\{c_n\},\{b_n\})\ge\varrho(\{a_n\},\{b_n\})
$$
$$
\Rightarrow \left(\sum_{n=0}^\infty{(a_n-c_n)^2}\right)^{1/2}+\left(\sum_{n=0}^\infty{(c_n-b_n)^2}\right)^{1/2}\ge\left(\sum_{n=0}^\infty{(a_n-b_n)^2}\right)^{1/2}
$$
Squaring both sides we get (from here on I will remove the summation bounds for clarity)
$$
\left(\sum{(a_n-c_n)^2}\right)+\left(\sum{(c_n-b_n)^2}\right)+2\sqrt{\left(\sum{(a_n-c_n)^2}\right)\left(\sum{(c_n-b_n)^2}\right)}\ge\left(\sum{(a_n-b_n)^2}\right) 
$$
The Cauchy-Schwarz Inequality gives us
$$
\left(\sum{(a_n-c_n)^2}\right)\left(\sum{(c_n-b_n)^2}\right)\ge\left(\sum{(a_n-c_n)(c_n-b_n)}\right)^2
$$
Taking the LHS and using the Cauchy-Schwarz Inequality, we have
$$
LHS\ge\left(\sum{(a_n-c_n)^2}\right)+\left(\sum{(c_n-b_n)^2}\right)+2\left(\sum{(a_n-c_n)(c_n-b_n)}\right)
$$
Expanding, we have
\begin{equation}
\begin{split}
LHS &\ge\left(\sum{(a_n^2+c_n^2-2a_n c_n)}\right)+\left(\sum{(b_n^2+c_n^2-2b_n c_n)}\right)+2\left(\sum{(a_n-c_n)(c_n-b_n)}\right) \\
&=\left(\sum{a_n^2}+\sum{c_n^2}-2\sum{a_n c_n}\right)+\left(\sum{b_n^2}+\sum{c_n^2}-2\sum{b_n c_n}\right) \\
&\ \ \ \ +2\left(\sum{(a_n-c_n)(c_n-b_n)}\right)
\end{split}
\end{equation}
By adding a nil factor $\left(2\sum{a_nb_n}-2\sum{a_nb_n}\right)=0$ we get
\begin{equation}
\begin{split}
LHS &\ge\sum{a_n^2}+\sum{c_n^2}-2\sum{a_n c_n}+\sum{b_n^2}+\sum{c_n^2}\\ 
&\ \ \ \ -2\sum{b_n c_n}+2\sum{a_nb_n}-2\sum{a_nb_n}+2\left(\sum{(a_n-c_n)(c_n-b_n)}\right)\\
&=\sum{(a_n^2+b_n^2-2a_nb_n)}+2\sum{(c_n^2-a_nc_n-b_nc_n+a_nb_n)}+2\sum{(-1)(c_n-a_n)(c_n-b_n)}\\
&=\sum{(a_n+b_n)^2}+2\sum{(c_n-a_n)(c_n-b_n)}-2\sum{(c_n-a_n)(c_n-b_n)}\\
&=\sum{(a_n+b_n)^2}=RHS
\end{split}
\end{equation}
Thus we have $LHS\ge RHS$ and the triangle inequality holds for $\varrho$.
Therefore, the four required conditions for a metric have been satisfied, and $\varrho$ is a metric on $\ell_2$.

What are your thoughts? My proof is certainly a mess, but is it correct?
 A: I think it is easier to think of this in terms of inner products. $l^2(\mathbb R)$ admits the natural inner product
$$\langle x,y\rangle = \sum_{i=0}^\infty x_iy_i.$$
This induces the norm
$$\|x\|_{l^2} = \left(\sum_{i=0}^\infty x_i^2\right)^{\frac12}.$$
To see that $\|\cdot\|_{l^2}$ is indeed a norm is straightforward; if $a\in\mathbb R$ and $x,y\in l^2$ then
$$\begin{align*}
\|ax\|_{l^2} = \left(\sum_{i=0}^\infty (ax_i)^2\right)^{\frac12} &= \left (a^2\sum_{i=0}^\infty x_i^2\right)^{\frac12}\\ &= |a|\left(\sum_{i=0}^\infty x_i^2\right)^{\frac12}\\
&= |a|\|x\|_{l^2}.  \end{align*}$$
Moreover,
$$\|0\|_{l^2} = \sum_{i=0}^\infty 0^2 = 0$$ and since $x_i^2>0$ iff $x_i\ne 0$, $\|x\|_{l^2} = 0$ implies $x=0$. For subadditivity, we have by the Minkowski inequality
$$\begin{align*}
\|x+y\|_{l^2} &= \left(\sum_{i=0}^\infty (x_i+y_i)^2\right)^{\frac12}\\
&\leqslant \left(\sum_{i=0}^\infty x_i^2\right)^{\frac12} + \left(\sum_{i=0}^\infty y_i^2\right)^{\frac12}\\
&= \|x\|_{l^2} + \|y\|_{l^2}.
\end{align*}$$
This norm induces a metric by 
$$\varrho(x,y) = \|x-y\|_{l^2}.$$
Clearly $\varrho$ is nonnegative since $\|\cdot\|_{l^2}$ is nonnegative, and $\|x-y\|_{l^2}=\|y-x\|_{l^2}$ since $(x_i-y_i)^2=(y_i-x_i)^2$. The triangle inequality follows from subadditivity of the norm:
$$\|x-y\|_{l^2} + \|y-z\|_{l^2} \leqslant \|(x-y)+(y-z)\|_{l^2} = \|x-z\|_{l^2} $$
