The initial investment by John is 50 USD. The daily interest earning is 2% of the current invested amount. Once the interest earning reaches 1 USD (which will take a day initially) John adds it back as an investment (so that the net investment after 1 day is 50 USD + 0.02(50 USD) = 51 USD). How much time will it take John to reach 100 USD?
I looked into how to calculate compound interest but all I am able to find out is ways of calculating when compounding is done at fixed intervals of time. Any guidance in the right direction would be much appreciated.
If John can collect simple interest and instantly reinvest it at any arbitrary instant, we can have a series of events like the following, where $y(t)$ is the balance in the account at time $t$:
At time $t_0 = 0,$ the account starts with balance $y(0) = 50.$
At time $t_1 = 1,$ the interest on the balance $50$ is $50 \times1(0.02)= 1.$ Reinvesting this, the balance becomes $y(1) = 51.$
Between times $t_1=1$ and $t_2=1+\frac{50}{51},$ the interest on the balance is $51 \times \frac{50}{51}(0.02) = 1.$ Reinvesting this, the balance becomes $y\left(1+ \frac{50}{51}\right) = 52.$
Between times $t_2=1+\frac{50}{51}$ and $t_3=1+\frac{50}{51}+\frac{50}{52},$ the interest on the balance is $52 \times \frac{50}{52}(0.02) = 1.$ Reinvesting this, the balance becomes $y\left(1+ \frac{50}{51}+\frac{50}{52}\right) = 53.$
This process ends after exactly $50$ reinvestments, since the amount reinvested each time is exactly $1$ and the goal is to increase the balance from $50$ to $100.$ The exact amount of time taken is $t$ such that $$ y(t) = y\left(1+ \frac{50}{51}+\frac{50}{52}+\ldots +\frac{50}{99}\right) = 100. $$
So literally the question is asking for the sum $$t_{50} = 1+ \frac{50}{51}+\frac{50}{52}+\ldots +\frac{50}{99}.$$ This can be tediously computed with a simple handheld calculator, or somewhat more easily by a programmable calculator or software spreadsheet.
The sum can also be approximated by solving for $t$ in the equation $$ e^{0.02t} = 2,\tag1 $$ where $e$ is the base of the natural logarithm. The solution to this is $t = 50\ln 2.$
But the solution for $t$ in Equation $1$ is also an approximation of the solution for $t$ in $$ 1.02^t = 2. $$ The solution is $$ t = \frac{\ln 2}{\ln 1.02}. $$ In fact, $\frac{\ln 2}{\ln 1.02}$ and $t_{50}$ are closer to each other than either of them is to $50\ln 2.$ So indeed if you want an approximate answer, it's better to solve for $t$ in $1.02^t = 2$ than in $e^{0.02t} = 2.$ The difference is $\frac{\ln 2}{\ln 1.02} - t_{50} \approx 0.1,$ that is, the procedure of reinvesting $1$ unit of value at gradually decreasing intervals of time reaches the goal in about $0.1$ time periods less than the procedure of reinvesting gradually increasing amounts of value at intervals of exactly $1$ time period, with one fractional time period at the end during which the balance reaches exactly $100.$
This close similarity to the result of conventional compound interest, along with the relative awkwardness of making sums like $t_{50}$ rather than simply taking powers of $1.02,$ helps explain why schemes like John's have not occurred much (if ever) in real life and why you will have such a difficult time to find them described anywhere.