Does $n^n = n! = (n+1)^n$ when computing a limit as n approaches infinity 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.
 A: 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 $+\infty$.
A: 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 $(n+x)^{(n+y)}\sim e^{x}n^n$
A: 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 $n \to \infty$.
A: 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.
