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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 $\lim_{n \to \infty}S_n=\infty$.

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    I think it's fine, but I'd rather begin by defining the function and then directly applying the logarithm to it. If the function diverges to infinity so do the values it takes on the naturals...2012-11-12
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    Why isn't $\lim_{x \to 3}f(x)= \infty$ for $x \in (-3,3)$ possible?2012-11-12
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    I've no idea what $\,f\,$ are you talking about...2012-11-12
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    Is there a proof for 'If the function diverges to infinity so do the values it takes on the naturals'?2012-11-12
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    In this case is (for example, the function you and I were talking about is continuous). It is based on the following trivial remark: for a continuous function $\,g\,$ , we have that $$\lim_{x\to x_0}g(x)=L\Longrightarrow \lim_{n\to\infty}g(x_n)=L$$ for any sequence s.t. $\,x_n\xrightarrow [n\to\infty]{} x_0\,$ within the domain of $\,g\,$.2012-11-12
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    The expression for $L(S(n))$ is wrong. If it were right, derivative would need interpretation.2012-11-12
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    @AndréNicolas:edited. Could you elaborate pls2012-11-12
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    You are saying somehow that the derivative of $\sum_1^n \log k$ is (in some sense) $\frac{1}{x}$. This is wrong. It is like arguing that $n^2=\sum_1^n n$ and therefore the derivative is $1$. Anyway, we can't differentiate the discrete function $n^2$.2012-11-12
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    I 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

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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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    Thanks. 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