Question #822e7

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Question #822e7

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Answer 1 · 2018-01-26T10:05:55

The series converges.

Explanation:

Trying to find $\infty \sum n = 1 (\frac{n^{n}}{n! \cdot 3^{n}})$ $sum_(n=1)^oo(n^n/(n!*3^n))$ (I'm staring at $n = 1$ $n=1$ instead of $n = 0$ $n=0$ because of the $0^{0}$ $0^0$ when $n = 0$ $n=0$ .)

Using the Ratio Test:
$lim n \to \infty ∣ ∣ ∣ ∣ \frac{{(n + 1)}^{n + 1}}{(n + 1)! \cdot 3^{n + 1}} \cdot \frac{n! \cdot 3^{n}}{n^{n}} ∣ ∣ ∣ ∣$ $lim_(n to oo)|(n+1)^(n+1)/((n+1)!*3^(n+1))*(n!*3^n)/n^n|$

$= lim n \to \infty ∣ ∣ ∣ \frac{(n + 1) {(n + 1)}^{n}}{(n + 1) \cdot n! \cdot 3^{n} \cdot 3} \cdot \frac{n! \cdot 3^{n}}{n^{n}} ∣ ∣ ∣$ $=lim_(n to oo)|((n+1)(n+1)^(n))/((n+1)*n!*3^(n)*3)*(n!*3^n)/n^n|$

$= lim n \to \infty ∣ ∣ ∣ \frac{{(n + 1)}^{n}}{3 n^{n}} ∣ ∣ ∣$ $=lim_(n to oo)|(n+1)^(n)/(3n^n)|$

$= \frac{1}{3} lim n \to \infty ∣ ∣ ∣ {(\frac{n + 1}{n})}^{n} ∣ ∣ ∣$ $=1/3lim_(n to oo)|((n+1)/n)^(n)|$

$= \frac{1}{3} lim n \to \infty {(1 + \frac{1}{n})}^{n}$ $=1/3lim_(n to oo)(1+1/n)^(n)$

$= \frac{1}{3} \cdot e$ $=1/3*e$ because $lim n \to \infty {(1 + \frac{1}{n})}^{n}$ $lim_(n to oo)(1+1/n)^(n)$ is famously equal to $e$ $e$ . [See this video for the

Since $\frac{e}{3} < 1$ $e/3<1$ , the series converges.

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Question #822e7

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