It's late, so I might have made an error somewhere.
\sum_{n=1}^{\infty} \frac{\frac{1}{4} + (-1)^n}{n}
= \sum_{n=1}^{\infty} \left(\frac{1}{4n} + \frac{(-1)^n}{n} \right)
= \frac{1}{4} \sum_{n=1}^{\infty} \left(\frac{1}{n} + \frac{(-1)^n}{n} \right)
= \frac{1}{4} \sum_{n=1}^{\infty} \frac{(-1)^n + 1}{n}
If n is equal =>
(-1)^n + 1 = 2.
If n is odd =>
(-1)^n + 1 = 0.
Therefore, we can write:
= \frac{1}{4} \sum_{n=1}^{\infty} \frac{2}{2n} = \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n}
Which is the harmonic series. The harmonic series is divergent.