  We prove that, given $\alpha>0$, if $M$ is a complete Riemannian manifold
which Ricci curvature
satisfies.\[\operatorname*{Ric}\nolimits_{x}(v)\geq\alpha\operatorname{sech}^{2}
(r(x)))\] or \[ \operatorname*{Ric}\nolimits_{x}(v)\geq-\frac{{h_{\alpha}}
(r(x))}{r(x)^{2}}, \] where \[ {h_{\alpha}}(r) =
\frac{\alpha(\alpha+1)r(x)^{\alpha }}{r(x)^{\alpha }-1}, \] for all $x\in
M\backslash B_{R}(o)$ and for all $v\in T_{x}M,$ $\left\Vert v\right\Vert =1,$
where \ $o$ is a fixed point of $M$, $r(x)=d(o,x)$, $d$ the Riemannian distance
in $M$ and $B_{R}(o)$ the geodesic ball of $M$ centered at $o$ with radius
$R>0$, then $M$ is $p-$parabolic for any $p>1$, if satisfies the first
inequality, and $M$ is $p-$parabolic, for any $p\geq(\alpha+1)(n-1)+1$, if
satisfies the second inequality.
