I was just thinking about math when a very simple thought occurred to me. Is there any way to express $0.39278124372921876561...$ as an infinite sum.
If you could not see the pattern in the number, it was basically all the powers of $3$ following the decimal point. So, expressing it as an sum of an infinite series from $n=0$ to infinity of $a(n)$ would be like:
$$a(0) = 0.3$$ $$a(1) = 0.09$$ $$a(2) = 0.0027$$ $$a(3) = 0.000081$$ $$a(4) = 0.000000243$$
...and so on.
So can you express this as a sum of an infinite series by basically putting the summation sign from $n=0$ to inf and then just putting an equation for $a(n)$. Or is that equation nonexistent?
Yes, you have the right idea. I'll use the more standard notation for sequences and write $a_{n}$ instead of $a(n)$. Then the number is precisely $$\sum_{n=0}^{\infty}a_{n}$$ The interesting part is coming up with a closed form for $a_{n}$. I'm not sure you'll be able to write this without a nested summation, but I've come up with the following: $$a_{n}=3^{n}\cdot 10^{-({\sum_{k=0}^{n-1}\lfloor \log_{10}3^{k}} \rfloor + 1)}$$ If it is unclear how I came up with this expression, think of it like this. The element $a_{n}$ is $3^{n}$ with some number of zeroes in front of it. The number of zeros corresponds to the number of digits we've already occupied by previous powers of $3$. So the number of zeros is the sum of the number of digits in all powers of $3$ before the one we are about to write. Since there are $\lfloor \log_{10}{N} \rfloor + 1$ digits in the number $N$, my expression follows.