Respuesta :
Answer:
Step-by-step explanation:
Recall that the ratio test is stated as follows:
Given a series of the form [tex]\sum_{n=1}^{\infty} a_n let L=\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|[/tex]
If L<1, then the series converge absolutely, if L>1, then the series diverge. If L fails to exist or L=1, then the test is inconclusive.
Consider the given series [tex]\sum_{n=1}^{\infty} n! \cdot 112n [/tex]. In this case, [tex]a_n =n! \cdot 112n[/tex], so , consider the limit
[tex]\lim_{n\to\infty} \frac{(n+1)! 112 (n+1)}{n! 112 n} = \lim_{n\to\infty}\frac{(n+1)^2}{n}[/tex]
Since the numerator has a greater exponent than the numerator, the limit is infinity, which is greater than one, hence, the series diverge by the ratio test
The series [tex]\Sigma\limits_{i=1}^{\infty} \left(\frac{n!}{112^{n}} \right)[/tex] is divergent.
How to determine the convergence of a series
In this question we must determine the convergence of a series, that is, if a series tends to a constant value when [tex]n \to +\infty[/tex]. The ratio test is a useful criterion for rational-like series:
[tex]r = \frac{a_{n+1}}{a_{n}}[/tex] (1)
Where the series converges if [tex]r < 1[/tex].
If we know that [tex]a_{n} = \frac{n!}{112^{n}}[/tex], then the ratio test is defined below:
[tex]r = \frac{\frac{(n+1)!}{112^{n+1}} }{\frac{n!}{112^{n}} }[/tex]
[tex]r = \frac{(n+1)!\cdot 112^{n}}{n!\cdot 112^{n+1}}[/tex]
[tex]r = \frac{(n+1)\cdot n!}{n!}\cdot \frac{112^{n}}{112\cdot 112^{n}}[/tex]
[tex]r = \frac{n+1}{112}[/tex]
[tex]r = \frac{n}{112} +\frac{1}{112}[/tex]
For [tex]n \to + \infty[/tex], [tex]r > 1[/tex], then the series is divergent. [tex]\blacksquare[/tex]
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