# Calc6.17

Discuss the convergence of $\sum _{{n=1}}^{{\infty }}{\frac {(-1)^{{n+1}}}{n^{2}+3n+4}}$ including whether the sum converges absolutely or conditionally.
$\sum _{{n=1}}^{{\infty }}\left|{\frac {(-1)^{{n+1}}}{n^{2}+3n+4}}\right|=\sum _{{n=1}}^{{\infty }}{\frac {1}{n^{2}+3n+4}}\,$
${\frac {1}{n^{2}+3n+4}}<{\frac {1}{n^{2}}}\,$
and we know that the series $\sum _{{n=1}}^{{\infty }}{\frac {1}{n^{2}}}$ converges. Thus, by the direct comparison test, the series $\sum _{{n=1}}^{{\infty }}\left|{\frac {(-1)^{{n+1}}}{n^{2}+3n+4}}\right|=\sum _{{n=1}}^{{\infty }}{\frac {1}{n^{2}+3n+4}}$ converges and thus our original series $\sum _{{n=1}}^{{\infty }}{\frac {(-1)^{{n+1}}}{n^{2}+3n+4}}$ not only converges but converges absolutely.