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.
Determine if the following series converges or diverges and justify your decision.
![\[ \sum_{n=1}^{\infty} \frac{(1000)^n}{n!}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-7a0edb4af5a045116d1ecf51caa610d7_l3.png "Rendered by QuickLaTeX.com")
The series converges by the ratio test since
![\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{1000^{n+1}}{(n+1)!} \right) \left( \frac{n!}{1000^n} \right) \\[9pt] &= \lim_{n \to \infty} \frac{1000}{n+1} \\[9pt] &= 0 < 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-d34b45425b75935cefe19b410ece1cc7_l3.png "Rendered by QuickLaTeX.com")
Hence, the series converges.
Determine if the following series converges or diverges and justify your decision.
![\[ \sum_{n=1}^{\infty} \frac{n!}{2^{2n}}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f752465ab32cd9d541b64d611adb5dbe_l3.png "Rendered by QuickLaTeX.com")
The series diverges by the ratio test since
![\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{(n+1)!}{2^{2n+2}} \right) \left( \frac{2^{2n}}{n!} \right) \\[9pt] &= \lim_{n \to \infty} \frac{n+1}{4} \\[9pt] &= + \infty > 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-82f20650f4eb5a8849c6ce0d85df5e7f_l3.png "Rendered by QuickLaTeX.com")
Hence, the series diverges.
Determine if the following series converges or diverges and justify your decision.
![\[ \sum_{n=1}^{\infty} \frac{n!}{3^n}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-9acd8ef98212e10702df6c3587aac563_l3.png "Rendered by QuickLaTeX.com")
The series diverges by the ratio test since
![\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{(n+1)!}{3^{n+1}} \right) \left( \frac{3^n}{n!} \right) \\[9pt] &= \lim_{n \to \infty} \frac{n+1}{3} \\[9pt] &= +\infty > 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-52b985f26dd0e64b415e0da086060b7e_l3.png "Rendered by QuickLaTeX.com")
Hence, the series diverges.
Determine if the following series converges or diverges and justify your decision.
![\[ \sum_{n=1}^{\infty} \frac{3^n n!}{n^n}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-747bd15b00313f28587e30510fdde39e_l3.png "Rendered by QuickLaTeX.com")
This series diverges by the ratio test since
![\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{3^{n+1}(n+1)!}{(n+1)^{n+1}} \right) \left( \frac{n^n}{3^n n!} \right) \\[9pt] &= 3 \lim_{n \to \infty} \left( \frac{n}{n+1}\right)^n \\[9pt] &= 3 \lim_{n \to \infty} \left( \frac{1}{1+\frac{1}{n}} \right)^n \\[9pt] &= \frac{3}{e} > 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-88090ee1ce515cf60a065d34db80c477_l3.png "Rendered by QuickLaTeX.com")
Hence, the series diverges.
Determine if the following series converges or diverges and justify your decision.
![\[ \sum_{n=1}^{\infty} \frac{2^n n!}{n^n}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-fffed4f12a172a9a62fdec691e1e1a49_l3.png "Rendered by QuickLaTeX.com")
The series converges by the ratio test since
![\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{2^{n+1}(n+1)!}{(n+1)^{n+1}} \right) \left( \frac{n^n}{2^n n!} \right) \\[9pt] &= \lim_{n \to \infty} \frac{2(n+1)n^n}{(n+1)^{n+1}} \\[9pt] &= \lim_{n \to \infty} \frac{2n^n}{(n+1)^n} \\[9pt] &= 2 \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^n \\[9pt] &= 2 \lim_{n \to \infty} \left( \frac{1}{1+\frac{1}{n}} \right)^n \\[9pt] &= \frac{2}{e} < 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-1b77adeb6f63ba2bac9b5b5eec7f9eaf_l3.png "Rendered by QuickLaTeX.com")
Hence, the series converges.
Determine if the following series converges or diverges and justify your decision.
![\[ \sum_{n=1}^{\infty} \frac{(n!)^2}{2^{n^2}}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-7cd93ae4013dd755aaa54062b1c1099c_l3.png "Rendered by QuickLaTeX.com")
The series converges by the ratio test since
![\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{(n+1)!^2}{2^{(n+1)^2}} \right) \left( \frac{2^{n^2}}{(n!)^2} \right) \\[9pt] &= \lim_{n \to \infty} \frac{(n+1)^2}{2^{2n+1}} \\[9pt] &= 0 < 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-73f6bfe1e512a928d3fec18e4f721934_l3.png "Rendered by QuickLaTeX.com")
Hence, the series converges.
Determine if the following series converges or diverges and justify your decision.
![\[ \sum_{n=1}^{\infty} \frac{(n!)^2}{(2n)!}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-4aab30652f95de2d614f8d265aa945a4_l3.png "Rendered by QuickLaTeX.com")
The series converges by the ratio test since
![\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{(n+1)!^2}{(2n+2)!} \right) \left( \frac{(2n)!}{(n!)^2} \right) \\[9pt] &= \frac{(n+1)^2}{(2n+1)(2n+2)} \\[9pt] &= \frac{n^2 + 2n + 1}{4n^2 + 6n + 2} \\[9pt] &= \frac{1}{4} < 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-483b5fc2861f274181167a6ae2d995ff_l3.png "Rendered by QuickLaTeX.com")
Hence, the series 
converges.