c) Suppose lim(n->inf) Sn = 0, this means that for all epsilon in R there exists
a natural number N such that:
|Sn - 0| < epsilon, for all n>N.
but this last is the same thing as:
|Sn| < epsilon, for all n>N
which is again the same as:
||Sn|-0| < epsilon, for all n>N
so lim(n->inf) Sn = 0 implies that lim(n->inf) |Sn| = 0.
For the other half of the proof, suppose lim(n->inf) |Sn| = 0, this means that
for all epsilon in R there exists a natural number N such that:
||Sn| - 0| < epsilon, for all n>N.
but this last is the same thing as:
|Sn| < epsilon, for all n>N
which is again the same as:
|Sn-0| < epsilon, for all n>N
so lim(n->inf) |Sn| = 0 implies that lim(n->inf) Sn = 0.
Together these proove: lim Sn = 0 iff lim |Sn| = 0
RonL