unit step signal as sum of impulses from MIT OCW video
yes there is only one term in the sum when $n \ge 0$
$$ \delta[m]=\begin{cases}1 \quad m=0 \\
0 \quad \text{otherwise} \end{cases}$$

$$ \begin{align}\sum_{m=-\infty}^n \delta[m] &amp;= \delta[-\infty]+ \dots +\delta[0] + \dots + \delta[n] \\
&amp; = 0 + \dots + 1 + \dots + 0  \\
&amp;= 1  \; \text{for} \; n\ge 0
\end{align}$$
The sum holds the value of 1 for $n\ge0$. His figure shows the terms inside the sum, not the sum of terms.

The definition of $u[n]$ is
$$ 
u[n]=\begin{cases} 1  \quad \text{for}\; n \ge 0 \\
0  \quad \text{for}\; n&lt;0 \end{cases} $$
They are equivalent.  Defining $u[n]$ function as a sum of delta functions may seem strange but I'm sure he has a "trick" to show that will be useful down the road.

There is a similar relationship in continuous time relating the unit step function $u(t)$ as the integral of the Dirac $\delta(t)$.