Answer:
The series converges to

Explanation:
It seems to be this series:

We have

Using the Root test we can see that this series converges once
![$ \lim_(n \to \infty) \sqrt[n]a_n < 1 \implies \sum_(n=1)^(\infty) a_n \text{ is convergent}$](https://img.qammunity.org/2022/formulas/mathematics/college/hoa386q8ln2z30b7ibpw26qfxe4sso3y2d.png)
Then,
![$\lim_(n \to \infty) \sqrt[n]{\left((1)/(4) \right)^(n)} = \lim_(n \to \infty) (1)/(4) = (1)/(4) < 1$](https://img.qammunity.org/2022/formulas/mathematics/college/bve5a4qqki25zre6oifig7bdyrpwnxo5rz.png)
The series is convergent.
Once the series is geometric, the first term is
and the ratio is also
in this case.
The sum of infinite geometric series is
such that

