# Is this line of thought for proving $\lim_{n \to \infty}S_n$ correct?

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... – DonAntonio Nov 12 '12 at 3:07
Why isn't $\lim_{x \to 3}f(x)= \infty$ for $x \in (-3,3)$ possible? – Alex Nov 12 '12 at 3:12
I've no idea what $\,f\,$ are you talking about... – DonAntonio Nov 12 '12 at 3:14
Is there a proof for 'If the function diverges to infinity so do the values it takes on the naturals'? – Alex Nov 12 '12 at 3:16
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 $\,g\,$. – DonAntonio Nov 12 '12 at 3:18

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. – Alex Nov 12 '12 at 21:53