The given series can be condensed to
[tex]\displaystyle \frac14 + \frac13 + \frac49 + \frac{16}{27} + \frac{64}{81} + \cdots = \frac14 \left(1 + \frac43 + \frac{4^2}{3^2} + \frac{4^3}{3^3} + \cdots\right) = \frac14 \sum_{n=0}^\infty \left(\frac43\right)^n[/tex]
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:
[tex]\displaystyle \lim_{n\to\infty} \frac14 \left(\frac43\right)^n = \frac14 \lim_{n\to\infty} \frac{4^n}{3^n} = \infty[/tex]
since 4ⁿ > 3ⁿ for all n > 0.
