Prove that running sum is equal to unit step function

Question

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.

Answer 1

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 $$ \sum_{k = 0}^{\infty} \delta[n-k] = \lim_{l \to \infty} a_l = 1 = u[n] \, . $$

Answer 2

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 $u[n]$.