# Calc6.18

Discuss the convergence of $\sum _{{n=1}}^{{\infty }}{\frac {(-1)^{n}}{n!}}$ including whether the sum converges absolutely or conditionally.
$\sum _{{n=1}}^{{\infty }}\left|{\frac {(-1)^{n}}{n!}}\right|=\sum _{{n=1}}^{{\infty }}{\frac {1}{n!}}$
For $n>3$, we have
${\frac {1}{n^{2}}}>{\frac {1}{n!}}$
and we know that $\sum _{{n=1}}^{{\infty }}{\frac {1}{n^{2}}}$ converges because it is a p-series with $p>1$. Thus, the series $\sum _{{n=1}}^{{\infty }}{\frac {1}{n!}}$ also converges by the direct comparison test. The first three terms do not change the convergence of a series. Adding a finite number to a convergent series does not change it to a divergent series. Finally, since this converges, we know that $\sum _{{n=1}}^{{\infty }}{\frac {(-1)^{n}}{n!}}$ converges absolutely.