Proof on Limits/Triangle Inequality
Let lim k→∞ ak=A. For all ε>0 , there exists an N∈N such that for all n1,n2≥N, |an1 −an2|<ε.
Pf: Let lim k→∞ ak=A. For all ε>0 , there exists an N∈N such that for all n1,n2≥N, |an1-A|<ε/2 and |an2-A|<ε/2. How would you use the triangle inequality at this point of the proof
proof-verification proof-explanation
asked Nov 29 at 19:07
Robert Klein
65
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Let lim k→∞ ak=A. For all ε>0 , there exists an N∈N such that for all n1,n2≥N, |an1 −an2|<ε.
Pf: Let lim k→∞ ak=A. For all ε>0 , there exists an N∈N such that for all n1,n2≥N, |an1-A|<ε/2 and |an2-A|<ε/2. How would you use the triangle inequality at this point of the proof
proof-verification proof-explanation
asked Nov 29 at 19:07
Robert Klein
65
add a comment |
Let lim k→∞ ak=A. For all ε>0 , there exists an N∈N such that for all n1,n2≥N, |an1 −an2|<ε.
Pf: Let lim k→∞ ak=A. For all ε>0 , there exists an N∈N such that for all n1,n2≥N, |an1-A|<ε/2 and |an2-A|<ε/2. How would you use the triangle inequality at this point of the proof
proof-verification proof-explanation
asked Nov 29 at 19:07
Robert Klein
65
Let lim k→∞ ak=A. For all ε>0 , there exists an N∈N such that for all n1,n2≥N, |an1 −an2|<ε.
Pf: Let lim k→∞ ak=A. For all ε>0 , there exists an N∈N such that for all n1,n2≥N, |an1-A|<ε/2 and |an2-A|<ε/2. How would you use the triangle inequality at this point of the proof
proof-verification proof-explanation
proof-verification proof-explanation
asked Nov 29 at 19:07
Robert Klein
65
asked Nov 29 at 19:07
Robert Klein
65
asked Nov 29 at 19:07
Robert Klein
65
asked Nov 29 at 19:07
Robert Klein
65
asked Nov 29 at 19:07
Robert Klein
65
65
add a comment |
add a comment |
1 Answer
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It's not clear to me whether your $n_1$ and $n_2$ are subscripts, but assuming they are, you have $$|a_{n_1}-a_{n_2}|=|(a_{n_1}-A)+(A-a_{n_2})|leq |a_{n_1}-A|+|A-a_{n_2}|< frac{1}{2}epsilon + frac{1}{2}epsilon.$$
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1 Answer
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1 Answer
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It's not clear to me whether your $n_1$ and $n_2$ are subscripts, but assuming they are, you have $$|a_{n_1}-a_{n_2}|=|(a_{n_1}-A)+(A-a_{n_2})|leq |a_{n_1}-A|+|A-a_{n_2}|< frac{1}{2}epsilon + frac{1}{2}epsilon.$$
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It's not clear to me whether your $n_1$ and $n_2$ are subscripts, but assuming they are, you have $$|a_{n_1}-a_{n_2}|=|(a_{n_1}-A)+(A-a_{n_2})|leq |a_{n_1}-A|+|A-a_{n_2}|< frac{1}{2}epsilon + frac{1}{2}epsilon.$$
add a comment |
It's not clear to me whether your $n_1$ and $n_2$ are subscripts, but assuming they are, you have $$|a_{n_1}-a_{n_2}|=|(a_{n_1}-A)+(A-a_{n_2})|leq |a_{n_1}-A|+|A-a_{n_2}|< frac{1}{2}epsilon + frac{1}{2}epsilon.$$
It's not clear to me whether your $n_1$ and $n_2$ are subscripts, but assuming they are, you have $$|a_{n_1}-a_{n_2}|=|(a_{n_1}-A)+(A-a_{n_2})|leq |a_{n_1}-A|+|A-a_{n_2}|< frac{1}{2}epsilon + frac{1}{2}epsilon.$$
answered Nov 29 at 19:52
user507152
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