It's not clear what your series is, so I'm going to take a wild guess on what it is you mean:
\(1-\dfrac{1\times3}{3!}+\dfrac{1\times3\times5}{5!}-\dfrac{1\times3\times5\times7}{7!}+\cdots\)
\(=\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n-1)!}\prod_{k=1}^n(2k-1)\)
For the sum to be absolute convergent, the sum of the absolute value of the summand must converge, so you are really examining the convergence of
\(\displaystyle\sum_{n=1}^\infty\frac1{(2n-1)!}\prod_{k=1}^n(2k-1)\)
This is easily checked with the ratio test:
\(\displaystyle\lim_{n\to\infty}\left|\frac{\displaystyle\dfrac1{(2(n+1)-1)!}\prod_{k=1}^{n+1}(2k-1)}{\displaystyle\dfrac1{(2n-1)!}\prod_{k=1}^n(2k-1)}\right|=\lim_{n\to\infty}\left|\frac{\dfrac{1\times3\times5\times\cdots\times(2n-1)\times(2n+1)}{(2n+1)!}}{\dfrac{1\times3\times5\times\cdots\times(2n-1)}{(2n-1)!}}\right|\)
\(\displaystyle=\lim_{n\to\infty}\left|\frac{\dfrac{2n+1}{(2n+1)(2n)}}{\dfrac11}\right|=\lim_{n\to\infty}\dfrac1{2n}=0<1\)
Since \(\sum|a_n|\) converges by the ratio test, the series \(\sum a_n\) converges absolutely.
