- #1
Math451
- 4
- 0
Let {Sn} be a sequence such that Sn > 0 for all n and Sn --> s , s > 0.
Prove that log (Sn) --> log (s) using the definition of convergence.
Also, we can use the following fact:
log(1 + x) <= x for all x > -1
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My attempt:
Point 1:
note that log(Sn) <= Sn - 1 and log(s) <= s - 1
which implies that
log(Sn) - log(s) <= (Sn-1) - (s -1),
Point 2:
Sn --> s means
for all epsilon, there exist N such that for all n > N implies
|Sn - s| < epsilon.
Thus, it follows from Point 1 and Point 2 that,
|log Sn| - |log s| <= |Sn-1| - |s-1| <= |(Sn-1) - (s - 1)| <= |Sn - s| < epsilon
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but |log Sn| - |log s| is not |log Sn - log s|.
so I am sort of stuck at this point.
Prove that log (Sn) --> log (s) using the definition of convergence.
Also, we can use the following fact:
log(1 + x) <= x for all x > -1
=====
My attempt:
Point 1:
note that log(Sn) <= Sn - 1 and log(s) <= s - 1
which implies that
log(Sn) - log(s) <= (Sn-1) - (s -1),
Point 2:
Sn --> s means
for all epsilon, there exist N such that for all n > N implies
|Sn - s| < epsilon.
Thus, it follows from Point 1 and Point 2 that,
|log Sn| - |log s| <= |Sn-1| - |s-1| <= |(Sn-1) - (s - 1)| <= |Sn - s| < epsilon
======
but |log Sn| - |log s| is not |log Sn - log s|.
so I am sort of stuck at this point.