how to execute ratio test for the following $\sum_{n=0}^\infty (-1)^n \frac{n!}{n^n}$
Wolfram says it converges by ratio test, however L = 1 when i execute the test. am I doing something wrong?
$L = \frac{n}{n+1}$
 A: Hint: Stirling's formula gives you
$$n\,!\sim \sqrt{2\pi n}\,\left({n \over {\rm e}}\right)^n$$

Alternately, if $a_n=(-1)^n\frac{n!}{n^n}$, then the ratio test runs like this:
$$\left|\frac {a_{n+1} }{a_n}\right|=  \frac{(n+1)!}{(n+1)^{n+1}} \frac{n^n}{n!}=(n+1) \cdot \left( \frac{n}{n+1}\right)^n \cdot \frac{1}{n+1}=\left( \frac{n}{n+1}\right)^n \underset{n \to \infty}\longrightarrow \frac{1}{e} < 1$$

For the last limit in the ratio test, I use the fact that $\left( 1 + \frac{1}{n}\right)^n \underset{n \to \infty}\longrightarrow e$. If you don't know this, you can prove it, using $\log \left(1+\frac{1}{n}\right) = \frac{1}{n} + O\left(\frac{1}{n^2}\right)$:
$$\left(1+ \frac{1}{n}\right)^n=\exp \left(n  \log \left(1+\frac{1}{n}\right)\right)=\exp \left(1+O\left(\frac{1}{n}\right)\right) \underset{n \to \infty}\longrightarrow e$$

If you don't know either this limit or the big-O notation, you can still do something: define
$$u_n=\left(1+\frac{1}{n}\right)^n$$
$$v_n=\left(1+\frac{1}{n}\right)^{n+1}$$
Then you can prove that $u_n$ is increasing, $v_n$ is decreasing (not difficult, but rather tedious algebraic manipulations), and of course $u_n < v_n$ since $v_n=\left(1+\frac{1}{n}\right) u_n$. Thus $u_n$ is convergent (this convergence follows from the fact that $u_n$ is increasing and bounded, by $v_1$), and its limit must be greater than $u_1=2$. Thus the limit in the ratio test is less than $1 \over 2$.
A: You could also consider the alternating series test:
$$\sum \limits_{i=1}^{\infty} (-1)^{n}a_{n}  $$ converges if the sequence $a_{n}$ is monotonically decreasing and null.
In your sum, the sequence satisfies this, so you do have convergence.
