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I am performing the ratio test on series and I come across many situations where I have any of those 3 in the numerator and denominator and I was wondering what I could cancel as n approaches infinity.

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  • $\begingroup$ You might look into the Stirling approximation. You'll find $ n^n $ and $(n+1)^n $ differ by a bounded factor while $ n! $ is far smaller. $\endgroup$
    – Ian
    Apr 4, 2015 at 16:22

4 Answers 4

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If I am interpreting your question correctly, here is how you can relate $n^n$ and $(n+1)^n$.

$$\lim_{n\to\infty} \frac{(n+1)^n}{n^n} = \lim_{n\to\infty} \left(\frac{n+1}{n}\right)^n = \lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^n = e.$$

More generally,

$$\lim_{n\to\infty} \frac{(n+k)^n}{n^n} = \lim_{n\to\infty} \left(\frac{n+k}{n}\right)^n = \lim_{n\to\infty} \left(1 + \frac{k}{n}\right)^n = e^k.$$

If you want to compare $n^n$ and $n!$ it might help to write $$\frac{n^n}{n!} = 1\cdot \frac{n}{n-1} \cdot \frac{n}{n-2} \cdots \frac{n}{2} \cdot \frac{n}{1} \geq n$$ which is a crude way to see that the limit must tend to $+\infty$.

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No, neither of these are true. First, $$ \lim_{n \to \infty} \frac{(n+1)^n}{n^n} = e. $$ This is because $$ \frac{(n+1)^n}{n^n} = \left(\frac{n+1}{n}\right)^n = \left(1 + \frac{1}{n}\right)^n $$ The limit of the last expression is the definition of $e$.

On the other hand $$ \lim_{n \to \infty} \frac{n!}{n^n} = 0. $$ That's because $$ \frac{n!}{n^n} = \frac{n \cdot (n-1) \cdot \ldots \cdot 1}{n \cdot n \cdot \ldots \cdot n} = \frac{n}{n} \cdot \frac{n}{n-1} \cdot \ldots \cdot \frac{1}{n}, $$ and that last product clearly goes to $0$ as $n \to \infty$.

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    $\begingroup$ And for $frac{n!}{(n+1)^n}$, not that $0<\frac{n!}{(n+1)^n} < \frac{n!}{n^n}$, so this one goes to $0$ as $n \rightarrow \infty$ as well. $\endgroup$
    – Roland
    Apr 4, 2015 at 16:28
  • $\begingroup$ what about $\frac{(n+1)^{k-1}}{n^k}$ where k is any number, I'm thinking that would be 0? $\endgroup$
    – Sean
    Apr 4, 2015 at 16:39
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We have the following orders we have $n!<<n^n$ and $(n+1)^n\sim e n^n$:

  • When we are dealing with fractions with theses terms the best approsimation for $n!$ is given by Stirling's formula: $$n\sim \left (\frac{n}{e} \right)^n\sqrt{2\pi n} $$
  • Note also that: $$\left (1+\frac{x}{n}\right)^n=e^x $$

so for example $(n+x)^{(n+y)}\sim e^{x}n^n$

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You should know the fundamental limit $$ \lim_{n\to +\infty} \frac{(n+1)^n}{n^n} = \lim_{n\to +\infty} \Bigl(\frac{n+1}{n}\Bigr)^n = \lim_{n\to +\infty} \Bigl(1 + \frac{1}{n}\Bigr)^n = e $$ so when $n$ approaches $+\infty$ the expressions $n^n$ and $(n+1)^n$ differ only by a constant. On the other hand we have $$ \lim_{n\to +\infty} \frac{n!}{n^n} = 0$$ and since we know the previous estimate, this limit gives also the relation between $(n+1)^n$ and $n!$. Another useful method is to use Stirling's approximation, which contains all the informations here explained.

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