determine whether the series ∑(10⋅20⋅30⋯(10k))/(10k)! converges or diverges. converges cannot be determined…

Explanation:

Step1: Apply the ratio - test

Let (a_{k}=\frac{10\cdot20\cdot30\cdots(10k)}{(10k)!}). Then (a_{k + 1}=\frac{10\cdot20\cdot30\cdots(10k)\cdot(10(k + 1))}{(10(k+1))!}). Calculate the ratio (\left|\frac{a_{k + 1}}{a_{k}}\right|=\frac{a_{k + 1}}{a_{k}}) (since (a_{k}>0) for (k\geq1)). [ \begin{align*} \frac{a_{k + 1}}{a_{k}}&=\frac{\frac{10\cdot20\cdot30\cdots(10k)\cdot(10(k + 1))}{(10(k + 1))!}}{\frac{10\cdot20\cdot30\cdots(10k)}{(10k)!}}\ &=\frac{10\cdot20\cdot30\cdots(10k)\cdot(10(k + 1))\cdot(10k)!}{10\cdot20\cdot30\cdots(10k)\cdot(10(k + 1))!}\ &=\frac{10(k + 1)}{(10k+1)(10k + 2)\cdots(10k+10)} \end{align*} ]

Step2: Find the limit as (k\to\infty)

We have (\lim_{k\to\infty}\left|\frac{a_{k + 1}}{a_{k}}\right|=\lim_{k\to\infty}\frac{10(k + 1)}{(10k+1)(10k + 2)\cdots(10k+10)}). The degree of the numerator is (1) and the degree of the denominator is (10). The leading - term of the numerator is (10k) and the leading - term of the denominator is ((10k)^{10}). (\lim_{k\to\infty}\frac{10(k + 1)}{(10k+1)(10k + 2)\cdots(10k+10)} = 0) since (\lim_{k\to\infty}\frac{10k+10}{(10k)^{10}}=0). Since (\lim_{k\to\infty}\left|\frac{a_{k + 1}}{a_{k}}\right| = 0<1), by the ratio - test, the series (\sum_{k = 1}^{\infty}\frac{10\cdot20\cdot30\cdots(10k)}{(10k)!}) converges.

Answer:

converges