I had exam from calculus $2$ and I could not determine convergence and absolute convergence of
$$\sum _{n=1}^{\infty }\frac{\left(-1\right)^n\left(1!+2!+...+n!\right)}{\left(n+1\right)!}$$
I tried to use linearity of sum and then use alternating series test. I am sure it is wrong. Same problem I had with absolute convergence test.
I would be grateful for any advice how to solve this problem.
$$ \frac{1!+...+n!}{(n+1)!}\geq\frac{1!+...+(n+1)!}{(n+2)!} \iff (1!+...+n!)(n+2) \geq 1!+...+(n+1)! \\ \iff (1!+...+n!)(n+1) \geq (n+1)! \iff 1!+...+n!\geq n! $$ And this is of course always true. This means that our sequence $a_n = \frac{1!+...+n!}{(n+1)!}$ is decreasing, and by Stolz theorem $$\lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{n!}{(n+1)!-n!} = \lim_{n\to\infty} \frac{1}{n} = 0.$$ From Leibniz test we see that the sum is convergent conditionally. It's not convergent absolutely since $a_n\geq \frac{1}{n+1} $.