How to see that the following series is convergent

Question

How can one check that the following series is convergent?

For any $\epsilon > 0$, does

$$\sum_{k=1}^{\infty} \left(\frac{1}{2^\epsilon}\right)^k$$

converge?

Answer 1

$$\sum_{k=1}^{\infty} \left(\frac{1}{2^\epsilon}\right)^k \le $$

$$\sum_{k=1}^{\infty} \left(r\right)^k = \frac {r}{1-r}$$

Where $0< r<1/{2^ {\epsilon }}<1$.

Answer 2

For the sake of having an answer:

$$ \sum_{k=1}^{\infty}\left(\frac{1}{2^{\epsilon}}\right)^{k}=\sum_{k=0}^{\infty}\left(\frac{1}{2^{\epsilon}}\right)^{k}-\left(\frac{1}{2^{\epsilon}}\right)^{0}=\frac{1}{1-\frac{1}{2^{\epsilon}}}-1=\frac{2^{\epsilon}}{2^{\epsilon}-1}-1=\frac{1}{2^{\epsilon}-1}. $$