Convergence of infinite series $\sum (-1)^{n+1}\frac{1}{n!}$ I have this question: Do the series converge absolutely or conditionally?
$$
\sum (-1)^{n+1}\frac{1}{n!}
$$
I would say it does not converge absolutely, since I suggest, by using the ratio test, that 
$$
\frac{a_{n+1}}{a_n} 
$$
does not approach a limit 
$$
L<1
$$
but rather 
$$
L=1
$$
However, I can see from the result list that I am wrong, and it does converge absolutely, can someone please explain to me why?
 A: The Ratio Test for absolute convergence should give you
$$\lim_{n \rightarrow \infty} \frac{1/(n+1)!}{1/n!} \ = \ \lim_{n \rightarrow \infty} \frac{n!}{(n+1)!} \ . $$
Be careful in setting up the ratio, since you have a compound fraction here...
EDIT: I might add that, after you've worked with infinite series for a while, you will develop a "feel" for what sorts of series are likely to be absolutely convergent.  We can make the comparison test  $  \frac{1}{n!} < \frac{1}{2^n} \ , $ for $ \ n > 3 \ $ , so the "infinite tail" of $ \ \Sigma_{n=1}^{\infty} \frac{1}{n!} \ $ is smaller than a convergent geometric series.  So it is reasonable to expect that this series should be absolutely convergent and "pass" the Ratio Test.
A: You do indeed have absolute convergence, and by the ratio test:
$$\lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to \infty} \frac{\frac{1}{(n+1)!}}{\frac 1{n!}} = \lim_{n\to \infty} \frac{n!}{(n+1)!} = \lim_{n\to \infty} \frac 1{n+1} = 0 < 1$$
Added: recall that the factorial is a product of factors, and so $$\frac{n!}{(n+1)!} = \frac{(n!)}{(n+1)(n!)} = \frac 1{n+1}$$
The factors given by $n! = n(n-1)(n-2)\cdots (2) \cdot (1)$ are common in both the numerator and denominator, and hence cancel, leaving only the factor of $1$ in the numerator, and $(n+1)$ in the denominator.
