An infinite matrix series Playing around with my CAS, I found that apparently
$$\sum_{n=1}^\infty\begin{bmatrix}1/2 & 1/3\\1/4 & 1/5\end{bmatrix}^n = \begin{bmatrix}29/19 & 20/19\\15/19 & 11/19\end{bmatrix}$$
If that is indeed true, it's pretty cool.
How can I prove this, and what is the general formula if the coefficients are replaced by arbitrary numbers $a,b,c,d<1$?

Given any $n \times n$ square matrix $A = (a_{ij})$ over a field $K = \mathbb{R}$ or $\mathbb{C}$. 
If its operator norm $\|A\|_{op}$:
$$\|A\|_{op} \stackrel{def}{=} \sup\big\{\; \|Av\| : v \in K^n \text{ with } \|v\| = 1 \;\big\}$$
or any sub-multiplicative matrix norm, e.g. the
Frobenius norm
$\|A\|_F$:
$$\|A\|_{F} \stackrel{def}{=} \sqrt{ \sum_{i=1}^n\sum_{j=1}^n |a_{ij}|^2 } = \sqrt{\text{tr}\left(A^{\dagger}A\right)}$$
is smaller than $1$, then the matrix series $\displaystyle\;\sum\limits_{k=1}^n A^n\;$
converge like what you find for a geometric series $\displaystyle\;\sum\limits_{k=1}^\infty z^k = \frac{z}{1-z}\;$ for $z \in K, |z| < 1$. More precisely,
$$\sum_{k=1}^\infty A^k = A\left(I_n - A\right)^{-1}\quad\text{ when }\quad\|A\| < 1 \tag{*1}$$
In particular, when $A = \begin{bmatrix}a & b\\c & d\end{bmatrix}$ and $\|A\|_F^2 = a^2 + b^2 + c^2 + d^2 < 1$, we have
$$\sum_{k=1}^\infty A^k 
= 
\begin{bmatrix}a & b\\c & d\end{bmatrix}
\begin{bmatrix}1-a & -b\\-c & 1-d\end{bmatrix}^{-1}
=
\frac{\begin{bmatrix}a - (ad-bc) & b\\ c & d - (ad-bc)\end{bmatrix}}{1-(a+d)+(ad-bc)}$$
In your case, $(a,b,c,d) = \left(\frac12,\frac13,\frac14,\frac15\right)$, its Frobenius norm $$\|A\| = \sqrt{\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2}} = \frac{\sqrt{1669}}{60} < 1$$
and the series converges to what you find
$$
\sum_{k=1}^\infty \begin{bmatrix}\frac12 & \frac13 \\ \frac14 & \frac15\end{bmatrix}^k = 
\frac{\begin{bmatrix}\frac12 - \frac{1}{60} & \frac13\\ \frac14 & \frac15 - \frac{1}{60}\end{bmatrix}}{1 - \frac{7}{10} + \frac{1}{60}} 
= \begin{bmatrix}\frac{29}{19} & \frac{20}{19}\\ \frac{15}{19} & \frac{11}{19}\end{bmatrix}$$
The proof of $(*1)$ is just like how you show $\sum_{k=1}^\infty z^k = \frac{z}{1-z}$ for $z \in K, |z| < 1$. You look at following partials sums and send $N$ to $\infty$
$$\require{cancel}
(1-z)\sum_{k=1}^N z^k = z - \cancelto{0 \text{ when } |z| < 1}{\color{grey}{z^{N+1}}}
\quad\leftrightarrow\quad
(I_n - A)\sum_{k=1}^N A^k = A - 
\cancelto{0 \text{ when } \|A\| < 1}{\color{grey}{A^{N+1}}}
$$
