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$$S=\sum^{\infty}_{n=1}\frac{n!}{n^n}z^n$$ Where $z \in \mathbb{C}$. Using D'Alambert's test of convergence:

$$\frac{1}{R}=\lim_{n\to \infty}\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}}$$ $$\frac{1}{R}=\lim_{n\to \infty}\frac{n^n}{(n+1)^n}$$

Where $R$ is the radius of convergence. When we consider the difference between $4^3$ and $3^3$, for example, it seems clear to me that this limit cannot be computed as follows:

$$\frac{1}{R}=\lim_{n\to \infty}\frac{n^n}{(n+1)^n}=1$$

However, this is exactly how my textbook (Riley, Hobson and Bence) does it. How can this computation be justified, considering that $n+1$ to a large power will always be a lot larger than $n$.

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2 Answers 2

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It is $\lim_{n\to\infty} \frac{n^n}{(n+1)^n}=\lim_{n\to\infty}\left(\frac{n}{1+n}\right)^n=\lim_{n\to\infty}\left(\frac{n+1-1}{n+1}\right)^n=\lim_{n\to\infty}\left(1-\frac{1}{n+1}\right)^n=\frac1e$

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  • $\begingroup$ Where did $e$ come from? $\endgroup$ Commented May 3, 2019 at 15:24
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    $\begingroup$ $ \lim_{n\rightarrow \infty}\frac{n^{n}}{(n+1)^{n}}=\lim_{n\rightarrow \infty}\left ( \frac{n}{n+1} \right )^{n}=\lim_{n\rightarrow \infty}\frac{1}{\left ( 1+\frac{1}{n} \right )^{n}}=\frac{1}{e} $ $\endgroup$
    – Student
    Commented May 3, 2019 at 15:25
  • $\begingroup$ @Pancake_Senpai It is the definition of $e^x$ as $\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n$. Note, that I had a typo in my calculation, which I edited by now. $\endgroup$
    – Cornman
    Commented May 3, 2019 at 15:26
  • $\begingroup$ In this case, it is $x=-1$. So we get $e^{-1}$ as the result. $\endgroup$
    – Cornman
    Commented May 3, 2019 at 15:27
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    $\begingroup$ @Pancake_Senpai That happens, but it is good, that you questioned the result. Many mistakes can be found, when you seriously study it and those minor detals are most of the time easy to fix. $\endgroup$
    – Cornman
    Commented May 3, 2019 at 15:40
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Hint: $$\frac{n^n}{(1+n)^n}=\frac{1}{(1+\frac{1}{n})^n}$$

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