Solution
= Solution
For every integer $n\geq1$, $\sin((2n-1)\pi)=0$ and $\cos((2n-3)\pi)=-1$. Thus the summand simplifies to $2/n$. The resulting <series> is twice the <harmonic series>, so \b[it diverges to $+\infty$].
= Solution
For every integer $n\geq1$, $\sin((2n-1)\pi)=0$ and $\cos((2n-3)\pi)=-1$. Thus the summand simplifies to $2/n$. The resulting <series> is twice the <harmonic series>, so \b[it diverges to $+\infty$].