Determine if the following series converges or diverges and justify your decision.
![\[ \sum_{n=1}^{\infty} \frac{n^3 \left( \sqrt{2} + (-1)^n \right)^n}{3^n}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-1eac29d0b4996433df3554cc6a1f8ea1_l3.png "Rendered by QuickLaTeX.com")
This series converges. First, we make the comparison,
![\[ \sum_{n=1}^{\infty} \frac{n^3 \left( \sqrt{2} + (-1)^n\right)^n}{3^n} \leq \sum_{n=1}^{\infty} \frac{n^3 \left( \sqrt{2} + 1\right)^n}{3^n}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-016938cc15688e2195369b4f1282b910_l3.png "Rendered by QuickLaTeX.com")
Then, for this series we use the ratio test, denoting the terms by 
we have,
![\begin{align*} \lim_{n \to \infty} \frac{b_{n+1}}{b_n} &= \lim_{n \to \infty} \left( \frac{(n+1)^3 \left( \sqrt{2} + 1 \right)^{n+1}}{3^{n+1}} \right) \left( \frac{ 3^n}{n^3 \left( \sqrt{2} + 1\right)^n } \right) \\[9pt] &= \lim_{n \to \infty} \frac{\left(\sqrt{2} + 1 \right) (n+1)^3}{3n^3} \\[9pt] &= \frac{\sqrt{2}+1}{3} < 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-7275e337ad12cacf199d27652f886b84_l3.png "Rendered by QuickLaTeX.com")
Hence, the series 
converges; therefore,
![\[ \sum_{n=1}^{\infty} \frac{n^3 \left( \sqrt{2} + (-1)^n \right)^n}{3^n} \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-5097526cd02d37204f118ee3131e75ab_l3.png "Rendered by QuickLaTeX.com")
converges as well.
You can also use the root test
the limit of the nth root does not exist, the terms keep oscillating so the root test fails
I think he might mean on the series we’re comparing to the original one (with a plus one instead of plus (-1)^n)