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Homework Statement
This question has been literally troubling me for 2 days.
Question 2(b)
Suppose for some c \in R, 0<c<1, we have |a_{n+1}-L|<c|a_{n}-L| for \forall n \in N (1)
Use the sandwich theorem and the fact that\stackrel{lim}{_{n\rightarrow\infty}} c^{n}=0 to prove that
\stackrel{lim}{_{n \rightarrow \infty}}a_{n}=L
Note: I also know and have proven in questions 2(a) that:
|a_{n}-L| \leq c^{n} |a_{0} - L| \forall n \in N(2)
Note:a_{n} is an element of the sequence (a_{n})
Homework Equations
Definition of a limit:
\forall n \in N, \forall \epsilon > 0, \exists K_{ \epsilon} \in R, n \ge K_{\epsilon} such that |a_{n}-L|< \epsilon
Sandwich theorem:
a_{n} \leq b_{n} \leq c_{n}, \stackrel{lim}{_{n \rightarrow \infty}}a_{n} = \stackrel{lim}{_{n \rightarrow \infty}}c_{n}=L \Rightarrow \stackrel{lim}{_{n \rightarrow \infty}}b_{n}=L
The Attempt at a Solution
for the sandwich theorem I could have b_{n} =c^{n} a_{n}. This is less than a_{n} since c is between zero and one. However the limit of this is 0(which is also under the assumption that the limit of a_{n} exists, which we are trying to prove. So starting off with sandwich theorem basics doesn't really help me because I also can't think of any c_{n} \geq a_{n}.
The next thing I tried is considering the statements in the problem.
|a_{n}-L| \leq c^{n} |a_{0} - L| could have been useful. But it doesn't imply anything about the sequence and its limits. I.E. I can't assume that this implies that a_{n} \leq c^{n}a_{0}
I can't really apply the definition of a limit to anything. By making the assumption that any of the equations above are less than some\epsilon almost automatically gives the result that the limit for a_{n} is L. i.e assuming the right hand side of equation (2) is less than some epsilon for \forall n>Kmeans|a_{n}-L|<\epsilon holds true as well.
Well it would almost hold true. But without knowing the details of a_{n} I caan't prove it to be true. I.E I can't construct a K_{\epsilon} unless I know more about a_{n}
PLEASE help on this one. I've been struggling for ages. Thanks
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