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Math Help - Use def of convergence

  1. #1
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    Use def of convergence

    Use the def of convergence to prove the sequence converges.

    Def: A sequence \{a_n\}_{n\in\mathbb{N}} converges to an A\in\mathbb{R} iif for each \epsilon >0 there exists N\in\mathbb{N} \ \text{such that} \ \forall n\geq N \ \text{we have} \ |a_n-A|<\epsilon

    \left\{5+\frac{1}{n}}\right\}_{n=1}^{\infty}

    So A = 5. Now, using the def.

    Let \epsilon >0 and there is N such that for n\geq N, \ \left | 5+\frac{1}{n}-5\right |=\left |\frac{1}{n}\right |<\epsilon

    \Rightarrow \frac{1}{n}\leq\frac{1}{N}<\epsilon

    \Rightarrow N>\frac{1}{\epsilon}

    Is that it?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Use def of convergence

    Quote Originally Posted by dwsmith View Post
    Use the def of convergence to prove the sequence converges.

    Def: A sequence \{a_n\}_{n\in\mathbb{N}} converges to an A\in\mathbb{R} iif for each \epsilon >0 there exists N\in\mathbb{N} \ \text{such that} \ \forall n\geq N \ \text{we have} \ |a_n-A|<\epsilon

    \left\{5+\frac{1}{n}}\right\}_{n=1}^{\infty}

    So A = 5. Now, using the def.

    Let \epsilon >0 and there is N such that for n\geq N, \ \left | 5+\frac{1}{n}-5\right |=\left |\frac{1}{n}\right |<\epsilon

    \Rightarrow \frac{1}{n}\leq\frac{1}{N}<\epsilon

    \Rightarrow N>\frac{1}{\epsilon}

    Is that it?
    Instead: We need to find such N.


    And... N is natural, but 1/epsilos is not always... so you can write N>[1/epsilon]+1 where [x] is the smallest integer than not greater than x.
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    Re: Use def of convergence

    The definition says one has to be able to find an N given an \epsilon. You showed that one can take any N>1/\epsilon, e.g., 1/\epsilon rounded up.

    You only need to change the direction of the arrows.

    N > 1/\epsilon and n > N \Rightarrow |a_n-A|=\frac{1}{n}\leq\frac{1}{N}<\epsilon
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    Re: Use def of convergence

    How would I handle:

    \{2^{-n}\}

    Would I say:

    N>\ln\left(\frac{1}{\epsilon}-2\right)\mbox{?}
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Use def of convergence

    2^{-n}=\frac{1}{2^n}


    And prove by induction :  2^n>n for n>2



     2^n>n or  \frac{1}{2^n}<\frac{1}{n}
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    Re: Use def of convergence

    Quote Originally Posted by Also sprach Zarathustra View Post
    2^{-n)=1/(2^n)
    I am aware of that.
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  7. #7
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    Re: Use def of convergence

    It should be any N>\ln\left(\frac{1}{\epsilon}\right)/\ln2=-\frac{\ln\epsilon}{\ln2}=-\log_2\epsilon.

    I am always trailing Zarathustra, but...

    You can also take any N>\frac{1}{\epsilon} because 2^n>n\Rightarrow 2^{-n}<1/n, so when n>N>\frac{1}{\epsilon}, we have 2^{-n}<1/n<1/N<\epsilon.
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  8. #8
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    Re: Use def of convergence

    one doesn't have to find "an optimal N" (although finding such an N can be evidence of a certain...cleverness), just one that's "big enough".

    in Zarathustra's example, the sequence {1/n} bounds {2^(-n)} from above, since {1/n} goes to 0 in the limit,

    and both sequences are positive, we have a special case of the "squeeze theorem" at work: {0} < {2^(-n)} < {1/n}.

    this is useful to remember when proving convergence: if the sequence is bounded by another sequence that makes the ε-algebra

    easier, go ahead and find an N for the "bigger sequence".

    of course the derivation of an optimal N is the case of 2^(-n) isn't that hard:

    we want: |2^(-n)| = 2^(-n) < ε. so log2(2^(-n)) < log2(ε) (log base 2 is increasing so preserves order)

    so -n < log2(ε), so we want n > -log2(ε) (the alternate form in emakarov's post no doubt comes from re-writing:

    2^(-n) = e^(-nln(2)), which has the advantage of being in the form of the standard log function, which makes it more

    transparent that ln is increasing, after all, it's derivative is 1/x, which is > 0 for all x in the domain of ln).
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