Solution
= Solution
For every integer $n$, $\cos((2n-1)\pi)=-1$. The <series> is therefore the negative of the <harmonic series>:
$$
\sum_{n=1}^N\frac{\cos((2n-1)\pi)}n=-\sum_{n=1}^N\frac1n.
$$
For a direct divergence argument, each block $2^{j-1}<n\le2^j$ contributes at least $2^{j-1}/2^j=1/2$ to the positive sum. Its <partial sums> grow without bound, so
$$
\boxed{\text{the given series diverges to }-\infty.}
$$
Despite the odd multiples of $\pi$, the signs do not alternate.