If the earth's rotational speed increased by 2% each day starting today…what would be the difference in age 20 years from now? If the new adjusted revolution of the earth still equaled one day and 365 days still equaled one year, how old would someone be 20 years from now (20 years based on the current rotation of the earth) compared to the new rotation of the earth?
I'm looking for a formula for the summation. (It is understood that a complete revolution around the sun would not be equal to 365 days. For the sake of the equation, leap year will not be factored in..1 year=365 days. It is also assumed that human life would still be possible given the earth's new rotational speed)
Edit 1:  The increase in speed will increase by 2% over the previous day's rotation, and it will happen at once at midnight.  Midnight tonight will indicate the beginning of the 20 year period (at the current rotational speed)
Edit 2:  I suppose a bit of suspended disbelief would be in order for this question. Someone who is adept in physics told me that human life and possibly the earth wouldn't even exist if the rotation of it increased at that amount for that length of time.  
 A: Since $speed \times time = distance$, and the distance is always constant, so on Day $1$, the speed is $1.02s$ ($s$ is the original speed), therefore the time needed for one full rotation is $t / 1.02$ ($t$ is the original time needed for one full rotation).
On Day $2$, the speed is $1.02^2s$. Therefore the time needed for one full rotation is $t / (1.02^2)$.
$\dots$
In general, on Day $x$, the speed is $1.02^xs$. The time needed for one full rotation is $t / (1.02^x)$.
Assume $20$ years is $7300$ days, now we need: 
$$\sum\limits_{x=1}^y \frac{t}{1.02^x} = 7300t$$
and solve for $y$. The $t$ on both sides can be cancelled out so we are left with:
$$\sum\limits_{x=1}^y \frac{1}{1.02^x} = 7300$$
Then we can apply the formula for the sum of the first $n$ terms of a geometric series and solve for $y$.
Added: Bad News! In fact there is no solution for $y$. Using sum of geometric series, we get:
$$\frac{1}{1.02} \frac{1-(\frac{1}{1.02})^{y+1}}{1-\frac{1}{1.02}} = 7300 $$
Simplifying gets:
$$ 50 \times (1-(\frac{1}{1.02})^{y+1}) = 7300 $$
As pointed out by @HagenvonEitzen, the LHS cannot exceed $50$ no matter how large $y$ is. So if my workings are correct, it suggests that the speed of the rotation goes so fast that in such a world, the infinity in terms of time is equal to $50$ days in the world we are now living in.
A: My original wrong answer:
$$
\frac{1}{365}\sum_{k=0}^{7299} 1.02^k\approx 8.278\cdot 10^{61}\text{ years}
$$

My original answer contained an error assuming that the increase followed the conventional days and not the new faster days. Here are my fixed versions:
Continuous version
If the speed increases gradually and by $2\%$ compared to the last midnight, it can be described via the following differential equation:
$$
y'=1.02^y
$$
where $y(x)$ denotes the number of rotations/"new days" after $x$ conventional days. Given the initial condition $y(0)=0$ this has solution
$$
y(x)=-\frac{\ln(1-\ln(1.02)x)}{\ln(1.02)}\approx -50.498\ln(1-0.0198026x)
$$
which has a vertical asymptote at $x=\frac{1}{\ln(1.02)}\approx 50.498$ so after approximately $50$ and a half conventional days, the earth reaches infinite rotational speed in this model.

Discrete version
The corresponding discrete model, similar to that given in LaBird's answer, should be
$$
x=\sum_{k=1}^y \frac{1}{1.02^{k-1}}=51\left(1-1.02^{-y}\right)
$$
which also has a vertical asymptote, this time at $x=51$.

Here is a graph of the two models (continuous as red curve, discrete as blue points):

A: Use $1$ real earth year as time unit, and denote the angular velocity of the earth rotation, measured in full turns per year, by $f(t)$. Then $f(0)=365$. When the angular velocity steadily increases by $2\%$ per day we have $f(t)=e^{\lambda t}f(0)$ for a certain $\lambda>0$, and this $\lambda$ is determined by the condition
$$e^{\lambda/365}=1.02\doteq e^{0.02}\ .$$
Neglecting the error here we obtain $\lambda=7.3$. Therefore the number of felt days during the next $20$ years from now is given by
$$\int_0^{20} f(t)\>dt=\int_0^{20} 365\>e^{7.3\>t}\>dt={365\over 7.3}\bigl(e^{146}-1\bigr)\ .$$
Counting the age as number of felt days divided by 365 produces a felt age of
$${1\over 7.3}\bigl(e^{146}-1\bigr)\doteq 3.49681\cdot10^{62}$$
years.
A: If I well understand the problem, after $1$ day the duration of a day become $d_1=\dfrac{1}{1+k}$ where $k=2\%$. So after $n$ days the duration af a day is $d_n=(1+k)^{-n}$, and the difference with respect the old days is $\Delta_n=n-\sum_{i=1}^n({1+k})^{-n}$. Using $n=365\times 20$ we have the result.
