Recent content by lamar333
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L
If a sequence converges to L, so the the sequence of averages
Yeah, I was just asked to prove this using nothing but the definition of convergence. As I struggled with it, I came across this thread and decided to post what I got.- lamar333
- Post #12
- Forum: Calculus and Beyond Homework Help
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L
If a sequence converges to L, so the the sequence of averages
Here's how I did it. (There are several ways, but I think this way is the most direct.) x_n-->L & y_n=(x_1+...+x_n)/n ==> y_n-->L PROOF: x_n-->L, so L-e<x_n<L+e for all n>N. So, x_n/(L-e)>1 and x_n/(L+e)<1. Now, let y_n =[x_1+...+x_(N-1)]/n + [x_N+...+x_n]/n. The first term...- lamar333
- Post #9
- Forum: Calculus and Beyond Homework Help