[Resolução] 10.16.4
Posted: 12 Oct 2022 17:16
\(\sum_{n=1}^{\infty} \frac{3^{n} n!}{n^n}\)
Iremos resolver usando o teorema da razão:
\(\lim_{n \to \infty} \frac{a_{n+1}}{a_{n}} = \lim_{n \to \infty} \frac{3^{n+1}(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{3^nn!}\)
\(\implies \lim_{n \to \infty} \frac{3n^n}{(n+1)^{n}} \implies 3\lim_{n \to \infty} \big(\frac{n}{n+1}\big)^n\)
\(\implies 3\lim_{n \to \infty} \big(\frac{1}{1+\frac{1}{n}}\big)^n \implies 3\lim_{n \to \infty} \big(\frac{1}{1+\frac{1}{n}}\big)^n = 3\frac{1}{e}\)
\(\implies \lim_{n \to \infty} \frac{a_{n+1}}{a_{n}} = \frac{3}{e}\)
Como \(\frac{3}{e} > 1\), devido ao teorema da razão, a série diverge.
Iremos resolver usando o teorema da razão:
\(\lim_{n \to \infty} \frac{a_{n+1}}{a_{n}} = \lim_{n \to \infty} \frac{3^{n+1}(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{3^nn!}\)
\(\implies \lim_{n \to \infty} \frac{3n^n}{(n+1)^{n}} \implies 3\lim_{n \to \infty} \big(\frac{n}{n+1}\big)^n\)
\(\implies 3\lim_{n \to \infty} \big(\frac{1}{1+\frac{1}{n}}\big)^n \implies 3\lim_{n \to \infty} \big(\frac{1}{1+\frac{1}{n}}\big)^n = 3\frac{1}{e}\)
\(\implies \lim_{n \to \infty} \frac{a_{n+1}}{a_{n}} = \frac{3}{e}\)
Como \(\frac{3}{e} > 1\), devido ao teorema da razão, a série diverge.