Recent content by dianetics
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If a sequence converges to L, so the the sequence of averages
I'm not following. Why do we only care about the terms of xn for sufficiently large n? yn is dependent on all x1...xn-1 as well. Edit: I have this so far: the convergence of x_n means that: | (x_n + x_n+1 + ... + xm) / (m - n + 1) - L | < e Now I'm trying to incorporate ym somehow into this...- dianetics
- Post #7
- Forum: Calculus and Beyond Homework Help
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If a sequence converges to L, so the the sequence of averages
How does this imply the convergence of y_n?- dianetics
- Post #6
- Forum: Calculus and Beyond Homework Help
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If a sequence converges to L, so the the sequence of averages
No, lol...- dianetics
- Post #4
- Forum: Calculus and Beyond Homework Help
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If a sequence converges to L, so the the sequence of averages
Homework Statement ( xn ) is given. ym = ( x1 + . . . + xm ) / m show that if xn converges to M, then ym converges to M Homework Equations Definition of convergence for sequences The Attempt at a Solution I've tried a whole page full of algebraic manipulation. I found a recursive...- dianetics
- Thread
- Sequence
- Replies: 11
- Forum: Calculus and Beyond Homework Help