\(\displaystyle{ a_n = \frac{n~\cos(n\pi)}{2n-1} }\)

\(\displaystyle{ \lim_{n \to \infty}{ \frac{n~\cos(n\pi)}{2n-1} } = \lim_{n \to \infty}{ \frac{\cos(n\pi)}{2-1/n} } }\)

In this limit, we know that \(\displaystyle{ \lim_{n \to \infty}{1/n} = 0 }\), leaving \(2\) in the denominator. In the numerator, \(\cos(n\pi)\) oscillates between \(-1\) and \(1\). So we are left with the series oscillating between \(-1/2\) and \(1/2\) and therefore, the limit \(\displaystyle{ \lim_{n \to \infty}{ \frac{n~\cos(n\pi)}{2n-1} } \neq 0 }\)
So, by the divergence test, the series diverges.

Conclusion
The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n~\cos(n\pi)}{2n-1} } }\) diverges by the divergence test.

If you look closely, you will notice that \(\cos(n\pi)\) is just a fancy way of writing \((-1)^{n}\). This is an alternating series which can be written \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n} n}{2n-1} } }\)
Let's try the alternating series test with \(\displaystyle{ a_n = \frac{n}{2n-1} }\)

Condition 1: We need to find the limit of \(\displaystyle{ \lim_{n \to \infty}{a_n} }\).
\(\displaystyle{ \lim_{n \to \infty}{a_n} = \lim_{n \to \infty}{ \frac{n}{2n-1} \frac{1/n}{1/n} } = \lim_{n \to \infty}{ \frac{1}{2-1/n} } = \frac{1}{2} \neq 0 }\)

Since the limit is not zero, condition 1 does not hold, so the alternating series test is inconclusive.