Suppose I want to calculate the number of bits it would take to store every (uncompressed) digital audio file up to 20 minutes in duration where each sample is stored with 32 bits and the sample rate is 44100 samples/second. Then is the expression for the number of bits given by:
$$32\times\sum_{t=1}^{20*60*44100}{t(2^{32})^t}$$
?
And, if so, what would be the order of magnitude of this number? Is there anything meaningful that it's comparable to?
In general,
$$ \sum_{t=1}^n t r^t = \frac{r (n r^{n + 1} - (n + 1) r^n + 1)}{(r - 1)^2}. $$
Plug in $n = 1200\times 44100$ and $r = 2^{32}$, then multiply the result by $32.$
For such a large $r,$ however, the numerator is dominated by the term $nr^{n+2}$ and the denominator is dominated by the term $r^2.$ You introduce a very small percentage error by ignoring all the other terms. This gives a total number of bits \begin{align} S \approx 32 nr^n &= 32 \times (1200\times 44100) \times 2^{32\times1200\times 44100} \\ &= 1693440000 \times 2^{1693440000}, \end{align} including the outermost factor $32.$
Equivalently, almost all (something like $99.9999\%$) of the space is occupied by files of the maximum length, so we can just count the space for files that are a full $20$ minutes long, so the length of each file is $L = 32 \times 1200\times 44100$ bits and there are $2^L$ of these files, resulting in the same approximation for the total size $S$ as given above.
Considering that the estimated number of protons, neutrons, and electrons in the observable universe is something on the order of $10^{80}\approx 2^{266},$ with maybe a few more orders of magnitude if we count all neutrinos, photons, and whatever is in dark matter, and it is evident that even if we could take every particle in the observable universe and replace it with an entire observable universe's worth of particles, the number of particles we would have would still be only a negligible fraction of the number of bits required.
This doesn't bode well for any kind of useful comparison to anything meaningful, except to say "it's bigger than that."