Answer
$\dfrac{4}{11}$
Work Step by Step
Re-write the given infinite series $\Sigma_{k=1}^{\infty} \dfrac{4}{12^k}$ as $\ 4 \
\Sigma_{k=1}^{\infty} \dfrac{1}{12^k}$
Here, we can see that the series $\Sigma_{k=1}^{\infty} \dfrac{1}{12^k}$ shows a geometric series with common ratio $r=\dfrac{1}{12}$ and $r=\dfrac{1}{12} \lt 1$. So, the series is convergent.
whose sum can be computed as: $\Sigma_{n=1}^{\infty} a^n =a+a^2+a^3+.....=\dfrac{a}{1-a}$
Now, $4 \Sigma_{k=1}^{\infty} \dfrac{1}{12^k}=4 \times \dfrac{\dfrac{1}{12}}{1-\dfrac{1}{12}}\\=4 \times \dfrac{11/12}{\dfrac{11}{12}}\\=\dfrac{4}{11}$
