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anonymous

  • 5 years ago

i)Prove, by using the formal definition of limit, that the sequence (1 + 5n)/n + n i / n + 3 n ∈ N has limit 5 + ı. ii)Prove that the fact that the above sequence converges to 5 + i implies that the sequence square root of ( [(1 + 5n)^2 / n^2] + [n^2 / (n+3)^2] ), n ∈ N converges to √26. Note: In this question you can assume without proof that the inequalities ||z| − |w|| ≤ |z + w| ≤ |z| + |w| hold for all z and w in C can anyone help me with this example? thanks

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  1. anonymous
    • 5 years ago
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    For the first part, start with the rearrangement,\[\frac{1+5n}{n}+\frac{n}{n+3}i=\frac{\frac{1}{n}+5}{1}+\frac{1}{1+\frac{1}{n}}i\]I don't know how fundamental an approach you have to take, but you can use the limit laws to write,\[\lim_{n->\infty}\left( \frac{\frac{1}{n}+5}{1}+\frac{1}{1+\frac{1}{n}} \right)\]\[=\frac{(\lim_{n->\infty}\frac{1}{n})+5}{1}+\frac{1}{1+(\lim_{n->\infty}\frac{1}{n})}\]Now choose \[\epsilon >0\]small. Then for \[\frac{1}{\epsilon}\]choose N such that N is the first integer greater than 1/(epsilon). Then \[\forall n>N ( n>\frac{1}{\epsilon} \iff \epsilon >\frac{1}{n}\iff \frac{1}{n}< \epsilon)\]Since \[n>0 \rightarrow \frac{1}{n}-0=\left| \frac{1}{n}-0 \right|\]and since \[\frac{1}{n}-0< \epsilon\]it follows,\[\left| \frac{1}{n} -0\right|< \epsilon\]Hence by the definition of the limit, \[\frac{1}{n} \rightarrow 0 \]as \[n \rightarrow \infty.\]

  2. anonymous
    • 5 years ago
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    Great, that worked.

  3. anonymous
    • 5 years ago
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    thats just taking limits tho isnt it. Thats not using the definition of the limit.

  4. anonymous
    • 5 years ago
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    The definition is used to prove 1/n goes to zero as n goes to infinity.

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