use the ratio test to determine whether the series is convergent or diverge. \n∑(n = 1 to ∞) (n^400 * 400^n)…

Explanation:

Step1: Identify (a_n)

Given the series (\sum_{n = 1}^{\infty}\frac{n^{400}400^{n}}{n!}), we have (a_n=\frac{n^{400}400^{n}}{n!}).

Step2: Find (a_{n + 1})

Replace (n) with (n+1) in (a_n). So (a_{n+1}=\frac{(n + 1)^{400}400^{n+1}}{(n+1)!}).

Step3: Calculate (\left|\frac{a_{n + 1}}{a_n}\right|)

[ \begin{align*} \left|\frac{a_{n + 1}}{a_n}\right|&=\left|\frac{\frac{(n + 1)^{400}400^{n+1}}{(n+1)!}}{\frac{n^{400}400^{n}}{n!}}\right|\ &=\left|\frac{(n + 1)^{400}400^{n+1}n!}{(n+1)!n^{400}400^{n}}\right|\ &=\left|\frac{(n + 1)^{400}}{n^{400}}\cdot\frac{400^{n+1}}{400^{n}}\cdot\frac{n!}{(n + 1)!}\right|\ &=\left(\frac{n + 1}{n}\right)^{400}\cdot400\cdot\frac{1}{n+1}\ &=\left(1+\frac{1}{n}\right)^{400}\cdot\frac{400}{n + 1} \end{align*} ]

Step4: Evaluate the limit (\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_n}\right|)

We know that (\lim_{n\rightarrow\infty}(1+\frac{1}{n})^{400}=1) and (\lim_{n\rightarrow\infty}\frac{400}{n + 1}=0). So (\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_n}\right|=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{400}\cdot\frac{400}{n + 1}=0).

Answer:

(a_n=\frac{n^{400}400^{n}}{n!}), (\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_n}\right| = 0), since (\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_n}\right|<1), the series is convergent.