The geometric series $\sum_{k=0}^{\infty}{ar^k}$ equals $\frac{a}{1-r}$ if and only if $% $.

By squinting at the notation, we can notice (not prove) a “well-known” matrix identity. (I keep forgetting it, so it’s not well-known to me).

If we let $a = 1$ and make the following replacements:

(where $A$ is some matrix and $I$ is the identity matrix), we can get the following “identity”:

Now we interpret “1 over something” as its multiplicative inverse, so the right hand side becomes $(I-A)^{-1}$.

Therefore, by our “identity”,

which is actually true.