Find the range of ppp for which ∑n=2∞1nplnn\sum_{n=2}^{\infty} \frac{1}{n^p \ln n}∑n=2∞nplnn1 converges.
p>0p > 0p>0
p>1p > 1p>1
p>1p > 1p>1 or (p=1p=1p=1 and q>1q > 1q>1)
p≥1p \geq 1p≥1