The given series can be condensed to
![\displaystyle \frac14 + \frac13 + \frac49 + (16)/(27) + (64)/(81) + \cdots = \frac14 \left(1 + \frac43 + (4^2)/(3^2) + (4^3)/(3^3) + \cdots\right) = \frac14 \sum_(n=0)^\infty \left(\frac43\right)^n](https://img.qammunity.org/2023/formulas/mathematics/high-school/adln3bdbrawjhyhx7oxa8l0wsmm8swcngi.png)
which is a geometric series with ratio 4/3. Since this ratio is larger than 1, the sequence of partial sums of the series diverges, so the infinite series also diverges.
We can also just the n-th term test:
![\displaystyle \lim_(n\to\infty) \frac14 \left(\frac43\right)^n = \frac14 \lim_(n\to\infty) (4^n)/(3^n) = \infty](https://img.qammunity.org/2023/formulas/mathematics/high-school/pp6rws01fre60y74ctotsbj1vdn67n6jnx.png)
since 4ⁿ > 3ⁿ for all n > 0.