Compounding by interest reaches a fixed value 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. 
 A: 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.
