Nested Geometric Series Formula I would like to derive a general formula for series of the following type:
$$\sum_{n=1}^{\infty}\left(\left(\frac{1}{A^n}\right) + \sum_{j=n}^{\infty}\left(\frac{1}{A^n\cdot B^j}\right)\right)$$
I attempted first to decompose it into parts by considering the nested loop:
$$n=1$$
$$\frac{1}{A}\sum_{j=1}^{\infty}\left(\frac{1}{B^j}\right)$$
which then allows for $n=1$ to be defined as:
$$\frac{1}{A} * \left(\frac{\frac{1}{B}}{1-\frac{1}{B}}\right)$$
then, letting $n$ iterate:
$$\frac{1}{A},\frac{1}{A^2},\frac{1}{A^3},...,\frac{1}{A^n},...$$
However, I'm having difficulty finding a rewrite for this nested loop. If one does exist, how would I extend that if I had a triply nested loop and so on. Thanks!

Let's ignore the outer summation for now and first just simplify the inner sum.
\begin{align}
\sum_{j=n}^{\infty} \frac{1}{A^n B^j} = \frac{1}{A^n}\sum_{j=n}^{\infty}\frac{1}{B^j} = \frac{1}{A^n B^n}\sum_{j=0}^\infty\frac{1}{B^j} = \frac{1}{A^n B^n}\frac{1}{1-1/B}
\end{align}
Notice that the power of $A$ doesn't depend on $j$, so the $A$ term can hop outside the sum. After that, we factored out the first term for convenience. Finally, we applied the standard geometric series formula with common ratio $1/B$.
Now let's get back to the outer sum
\begin{align}
\sum_{n=1}^\infty \frac{1}{A^n} + \frac{1}{(AB)^n}\frac{1}{1-1/B} 
= \sum_{n=1}^\infty \frac{1}{A^n} + \frac{1}{1-1/B}\sum_{n=1}^\infty\frac{1}{(AB)^n}
\end{align}
We've split the sum over the two terms and we took the $1/(1-1/B)$ out of the summation because it doesn't depend on $n$. I think from this point, you will be able to finish the problem.

However, I'm having difficulty finding a rewrite for this nested loop. If one does exist, how would I extend that if I had a triply nested loop and so on.

In this case, as you can see it is easier to not work with individual terms but to keep the indices (instead of plugging them in immediately) and try to simplify. Plugging in indices, then writing the sequence, and then creating indices again by spotting the pattern is more prone to error and I would avoid it if possible.
