What does the p-series test say about ∑n=1∞1nπ\sum_{n=1}^{\infty} \frac{1}{n^{\pi}}∑n=1∞nπ1?
It converges since p=π>1p = \pi > 1p=π>1
It diverges since p=π>1p = \pi > 1p=π>1
It converges since p=π<1p = \pi < 1p=π<1
It diverges since p=π<1p = \pi < 1p=π<1