## 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: $\sum_{n=1}^{\infty}\frac{1}{n^{\sqrt 2}}$ The given series is a p-series with $p= \sqrt 2 \gt 1$ and it is convergent. Hence, the series $\sum_{n=1}^{\infty}\frac{1}{n^{\sqrt 2}}$ is convergent.
The p-series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent if $p\gt 1$ and divergent if $p\leq 1$. Given: $\sum_{n=1}^{\infty}\frac{1}{n^{\sqrt 2}}$ The given series is a p-series with $p= \sqrt 2 \gt 1$ and it is convergent. Hence, the series $\sum_{n=1}^{\infty}\frac{1}{n^{\sqrt 2}}$ is convergent.