## Calculus 8th Edition

The p-series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent if $p\gt 1$ and divergent if $p\leq 1$. Given: $\sum_{n=3}^{\infty}n^{-0.9999}=\sum_{n=3}^{\infty}\frac{1}{n^{0.9999}}$ The series $\sum_{n=1}^{\infty}\frac{1}{n^{0.9999}}$is a p-series with $p= 0.9999 \leq 1$ and it is divergent. But the given series is not exactly a p-series, it is because the summation starts from $n=3$ But adding or subtracting finite number of terms will not change the convergence/divergence nature of the series. Hence, the given series is divergent.