Answer
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.
Work Step by Step
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.
