Famous businessman mathematical puzzle ${}{}$

The way I interpret this problem, the salaries go as follows:

Person 1:

He gets paid $2000$ for the first year, $2300$ for the second year, $2600$ for the third year, etc. So for the $n$th year he gets paid $2000+300(n-1)$.

Person 2:

He gets paid $1000+1100$ for the first year, $1200+1300$ for the second year, $1400+1500$ for the third year, etc. So he does end up getting paid more, year-by-year, because his raises occur more frequently and so can build up more.

Suppose if they get $100$ each half year. Then the following are the possible outcomes $$1^{st}\mbox{ year }\ \ \ \ \ \ 1000+1100=2100$$ $$2^{nd}\mbox{ year }\ \ \ \ \ \ 1200+1300=2500$$ $$3^{rd}\mbox{ year }\ \ \ \ \ \ 1400+1500=2900$$ $$4^{th}\mbox{ year }\ \ \ \ \ \ 1600+1700=3300$$

Suppose if they get $300$ each per year. Then the following are the possible outcomes $$1^{st}\mbox{ year }\ \ \ \ \ \ 1000+1000=2000$$ $$2^{nd}\mbox{ year }\ \ \ \ \ \ 1150+1150=2300$$ $$3^{rd}\mbox{ year }\ \ \ \ \ \ 1300+1300=2600$$ $$4^{th}\mbox{ year }\ \ \ \ \ \ 1450+1450=2900$$

Now can you see which one is profitable.

You are paid at the end of the work period, so in the first year the first man gets $1000$ twice for $2000$, then is raised to $2300/$year for the second year. The second gets $1000$ for the first six months and $1100$ for the second, giving a total of $2100$. He is ahead by $100$ after the first year.

The second year the first man gets $1150$ each time for a total of $2300$. The second gets another raise to $1200$ for the first six months and one to $1300$ for the second six months, giving a total of $2500$. He is ahead by $200$ in the second year.

In general, in year $k$, the first man gets $2000+300(k-1)$. The second gets $2000+2\cdot 100\cdot (k-1)+2\cdot 100 \cdot (k-1)+100=2000+400(k-1)+100$ and his advantage grows by $100$ each year.