Say I have $S_n = \frac{n^n}{n!}$ nad I want to show that $\lim_{n \to \infty}S_n= \infty$. Is the following line of though correct, and if not, where any why am I wrong? Here $L(\cdot) = \log(\cdot)$ and $n$ is replaced by $x$. $ S(n)=\frac{n^n}{n!}\\ L(S(n))=n \log n -\sum_{k=1}^{n}\log k\\ L'(S(x))=\log x+1+O\bigg(\frac{1}{x}\bigg)\\ \lim_{x \to \infty}L'(S(x))=\infty $ Hence $\lim_{n \to \infty}S_n=\infty$.
Is this line of thought for proving $\lim_{n \to \infty}S_n$ correct?
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limits
asymptotics
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0I see. I were trying to go about interchanging limit and logging and limit and differentiation. Could you maybe suggest a different problem where this would make sense. – 2012-11-12
1 Answers
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Here is an idea for you from infinite series. Let us look at the series
$\sum_{n=1}^\infty\frac{n!}{n^n}$
Applying D'Alembert's test, we get:
$\frac{(n+1)!}{(n+1)^{n+1}}\frac{n^n}{n!}=\frac{1}{\left(1+\frac{1}{n}\right)^n}\xrightarrow [n\to\infty]{} e^{-1}<1$
and thus the series converges, from where
$\frac{n!}{n^n}\xrightarrow [n\to\infty]{} 0\Longrightarrow \frac{n^n}{n!}\xrightarrow [n\to\infty]{}\infty$
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0Thanks. I know how to prove it. I were more interested in the correctness of the proof approach I used in the post. – 2012-11-12