Prove that running sum is equal to unit step function I'm looking for a formal proof of following equality:

$$ u[n] = \sum_{k = 0}^{\infty} \delta[n-k]$$

My try: We should prove that  $\lim_{k \to \infty}(\delta[n] + \delta[n-1] + \dots \delta[n-k]) = u[n]$. According to definitions $$u[n] =
\begin{cases}
0  & \text{if $n \lt0$} \\
1 & \text{if $n\ge0$ }
\end{cases}$$
and 
$$\delta[n] =
\begin{cases}
0  & \text{if $n \not=0$} \\
1 & \text{if $n =0$ }
\end{cases}$$
I don't know how to relate these equations to prove that equality. Intuitively the equality is obvious.
 A: Let $a_l = \sum_{k = 0}^{l} \delta[n-k]$ for $l = 0, 1, 2, \ldots$.
Case 1: $n < 0$. Then $n-k < 0$ for all $k \ge 0$, so that $a_l = 0$ for all $l \ge 0$. It follows that
$$
 \sum_{k = 0}^{\infty} \delta[n-k] = \lim_{l \to \infty} a_l = 0 = u[n] \, .
$$
Case 2: $n \ge 0$. For all $l \ge n$
$$
a_l = \sum_{k = 0}^{l} \delta[n-k] = \sum_{k = n}^{n} \delta[n-k] = 1
$$
because all other terms vanish. It follows that
$$
 \sum_{k = 0}^{\infty} \delta[n-k] = \lim_{l \to \infty} a_l = 1 = u[n] \, .
$$
A: It is simpler if you use the definitions. The Kronecker delta is
$$
\delta[n]=
\left\{
\begin{array}{r,l}
1, & n=0\\
0, & n\neq 0
\end{array}\right.
\quad\Rightarrow\quad
\delta[n-k]=
\left\{
\begin{array}{r,l}
1, & n-k=0\\
0, & n-k\neq 0
\end{array}\right.
\quad\Rightarrow\quad
\delta[n-k]=
\left\{
\begin{array}{r,l}
1, & n=k\\
0, & n\neq k
\end{array}\right..
$$
Since $k\geq0$ in the summation, it becomes
$$\sum_{k=0}^{\infty}\delta[n-k]=
\left\{
\begin{array}{r,l}
1, & n\geq 0\\
0, & n<0
\end{array}\right.,
$$
which in turn is the definition of the step function $u[n]$.
