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

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

    So A = 5. Now, using the def.

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

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

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

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

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

    So A = 5. Now, using the def.

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

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

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

    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 $\displaystyle N$ given an $\displaystyle \epsilon$. You showed that one can take any $\displaystyle N>1/\epsilon$, e.g., $\displaystyle 1/\epsilon$ rounded up.

    You only need to change the direction of the arrows.

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

    How would I handle:

    $\displaystyle \{2^{-n}\}$

    Would I say:

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

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


    And prove by induction :$\displaystyle 2^n>n$ for $\displaystyle n>2$



    $\displaystyle 2^n>n$ or$\displaystyle \frac{1}{2^n}<\frac{1}{n}$
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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 $\displaystyle N>\ln\left(\frac{1}{\epsilon}\right)/\ln2=-\frac{\ln\epsilon}{\ln2}=-\log_2\epsilon$.

    I am always trailing Zarathustra, but...

    You can also take any $\displaystyle N>\frac{1}{\epsilon}$ because $\displaystyle 2^n>n\Rightarrow 2^{-n}<1/n$, so when $\displaystyle n>N>\frac{1}{\epsilon}$, we have $\displaystyle 2^{-n}<1/n<1/N<\epsilon$.
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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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