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Test if the given series converges or diverges

Determine if the following series converges or diverges and justify your decision.

    \[ \sum_{n=1}^{\infty} \frac{n^3 \left( \sqrt{2} + (-1)^n \right)^n}{3^n}. \]


This series converges. First, we make the comparison,

    \[ \sum_{n=1}^{\infty} \frac{n^3 \left( \sqrt{2} + (-1)^n\right)^n}{3^n} \leq \sum_{n=1}^{\infty} \frac{n^3 \left( \sqrt{2} + 1\right)^n}{3^n}. \]

Then, for this series we use the ratio test, denoting the terms by b_n 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*}

Hence, the series \sum b_n converges; therefore,

    \[ \sum_{n=1}^{\infty} \frac{n^3 \left( \sqrt{2} + (-1)^n \right)^n}{3^n} \]

converges as well.

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