The video link doesn't work for me, but I assume this is the problem involving choosing between two payment types, one that immediately grants you some lump sum of cash, while the other starts off with $0.01 on the first day, $0.02 on the second day, $0.04 on the third, and so on, doubling per day.
Let [tex]m_n[/tex] be the amount of money earned on the [tex]n[/tex]-th day. Then [tex]m_n[/tex] is a geometric sequence that satisfies
[tex]\begin{cases}m_1=0.01\\m_n=2m_{n-1}&\text{for }n>1\end{cases}[/tex]
We can solve explicitly for [tex]m_n[/tex] in terms of the starting payment [tex]m_1[/tex]:
[tex]m_n=2m_{n-1}[/tex]
[tex]m_{n-1}=2m_{n-2}\implies m_n=2^2m_{n-2}[/tex]
[tex]m_{n-2}=2m_{n-3}\implies m_n=2^3m_{n-3}[/tex]
[tex]\cdots[/tex]
[tex]m_2=2m_1\implies m_n=2^{n-1}m_1[/tex]
So the amount of money earned on the [tex]n[/tex]-th day is
[tex]m_n=2^{n-1}\cdot0.01[/tex]
Denote the total amount of money earned over [tex]n[/tex] days by [tex]M_n[/tex]. Then we can write
[tex]M_n=\displaystyle\sum_{i=1}^nm_i=m_1+m_2+\cdots+m_{n-1}+m_n[/tex]
but since we have equivalent expressions for [tex]m_i[/tex] on any given day [tex]i[/tex], this is the same as
[tex]M_n=\displaystyle\sum_{i=1}^nm_i=0.01+2\cdot0.01+\cdots+2^{n-2}\cdot0.01+2^{n-1}\cdot0.01[/tex]
[tex]M_n=0.01(1+2+\cdots+2^{n-2}+2^{n-1})[/tex]
Let's call the sum on the right hand side [tex]S_n[/tex]. Notice that
[tex]S_n=1+2+\cdots+2^{n-2}+2^{n-1}[/tex]
[tex]2S_n=2+2^2+\cdots+2^{n-1}+2^n[/tex]
[tex]\implies S_n-2S_n=1-2^n[/tex]
[tex]\implies-S_n=1-2^n[/tex]
[tex]\implies S_n=2^n-1[/tex]
which means we end up with
[tex]M_n=0.01(2^n-1)[/tex]
To answer part (D), you need to look no further than the formula for [tex]M_n[/tex]. The question is basically asking what happens as [tex]n[/tex] gets arbitrarily large. It should be clear that [tex]2^n[/tex] grows without bound, so as [tex]n\to\infty[/tex], the amount of money you would get would diverge to infinity. So technically, you cannot calculate the amount because (1) there's only a finite amount of money to go around, and (2) infinity is not a "computable" number.
If, however, the scale factor used on your income was smaller than 1 - for example, say you were started with a million dollars on the first day, then your income got halved each day - then [tex]m_n[/tex] would eventually converge to 0, and on top of that, [tex]M_n[/tex] would converge to a finite number.
