Thread: 2 Proof of Sequences

1) Prove that lim (Sn) = 0 iff lim|(Sn)| = 0.
Note: the || are absolute values.

2) Let (Sn) be a convergent sequence, and suppose that lim Sn > a.
Prove that htere exists a number N such that n > N implies Sn > a.

These are to be proved using the definition the limit of a sequence:
For each epsilon > 0 there exists a number N such that
n > N implies |Sn - L| < epsilon

Thank you so much I am very stuck right now.

Originally Posted by tbyou87

1) Prove that lim (Sn) = 0 iff lim|(Sn)| = 0.
Note: the || are absolute values.

If lim s_n =0 that means,
|s_n|<e for n>N

Hence if you chose n>N we have,
||s_n|-0|=||s_n||=|s_n|<e
Thus,
lim |s_n| = 0
The same the other way around.

2) Let (Sn) be a convergent sequence, and suppose that lim Sn > a.
Prove that htere exists a number N such that n > N implies Sn > a.
.

Let lim s_n = L

By hypothesis L>a thus, L-a>0 thus (L-a)/2 >0

That means,

|s_n-L| < (L-a)/2 for n>N
Open absolute values,
-(L-a)/2< s_n - L < (L-a)/2
Thus,
L-(L-a)/2 < s_n < L + (L-a)/2
Thus,
(L+a)/2 < s_n < (2L-a)/2
The fist inequality says,
(L+a)/2 < s_n
Thus,
(a+a)/2 < (L+a)/2 < s_n for n>N
Thus,
a < s_n for all n>N