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Thread: Limit Theorems, Sequences

  1. #1
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    Limit Theorems, Sequences

    Suppose that (sn) and (tn) are convergent sequences with lim sn = s and lim tn = t. Then lim (sntn) = st.

    Write a proof to this theorem that does not use the theorem that states that "every convergent sequence is bounded." I think that this can be done by using the identity sntn - st = (sn - s)(tn - t) + s(tn - t) + t(sn - s). I am lost from here.
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  2. #2
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    If $\displaystyle \varepsilon > 0$ then choose $\displaystyle \delta = \min \left\{ {\sqrt {\frac{\varepsilon }{3}} ,\frac{\varepsilon }
    {{3\left( {\left| s \right| + 1} \right)}},\frac{\varepsilon }
    {{3\left( {\left| t \right| + 1} \right)}}} \right\}$.

    Make $\displaystyle \left| {s_n - s} \right| < \delta \;\& \,\left| {t_n - t} \right| < \delta $.

    Then use the identity.
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  3. #3
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    I'm still lost...
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    Quote Originally Posted by bearej50 View Post
    I'm still lost...

    The identities that you must use are:


    1)$\displaystyle s_{n}t_{n}-st = s_{n}t_{n}-st_{n}+st_{n}-st = t_{n}(s_{n}-s)+s(t_{n}-t)$

    OR

    2) $\displaystyle s_{n}t_{n}-ts_{n}+ts_{n}-st = s_{n}(t_{n}-t)+t(s_{n}-s)$.

    In using for example the (1) identity we can work as follows:

    we want to find out if there exists a No ,k, belonging to the natural Nos,such that for any given positive No ε>0

    $\displaystyle |s_{n}t_{n}-st| = |s_{n}t_{n}-st_{n}+st_{n}-st| = |t_{n}(s_{n}-s)+s(t_{n}-t)|<\epsilon$,whenever ,$\displaystyle n\geq k$.

    WHERE n is any natural No.

    But since $\displaystyle |t_{n}(s_{n}-s)+s(t_{n}-t)|\leq |t_{n}||s_{n}-s| + |s||t_{n}-t|\leq |t_{n}||s_{n}-s| + (|s|+1)|t_{n}-t|$,if we can find a ,k, such that for :

    $\displaystyle n\geq k$ ,then $\displaystyle |t_{n}||s_{n}-s|<\frac{\epsilon}{2}$ ,and

    $\displaystyle (|s|+1)|t_{n}-t|<\frac{\epsilon}{2}$,we will have proved the desired result.

    Note instead of using |s| we use ||s|+1| in the above inequalities is to avoid the case where s=0.

    HINT: In trying to find this ,k, it must be the max{$\displaystyle n_{1},n_{2},n_{3}$} i,e the maximum of three other natural Nos
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  5. #5
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    bearej50.

    Since there is no respond from you ,i will assume that you still do not know how to do the limit,so i will prove it.

    So we want to prove $\displaystyle \lim_{n\rightarrow\infty}{s_{n}t_{n}} = st$

    Let ε>0.

    Thus $\displaystyle \frac{\epsilon}{2(|t|+\epsilon)}>0$.................................................. ................................1


    and $\displaystyle \frac{\epsilon}{2(|s| +1)}>0$.................................................. ........................................2

    Since now $\displaystyle \lim_{n\rightarrow\infty}{s_{n}} = s$ and $\displaystyle \lim_{n\rightarrow\infty}{t_{n}} = t$,given any +ve No and thus even the +ve Nos (1),(2),ε,
    we can find Natural Nos $\displaystyle n_{1},n_{2},n_{3}$ and such that:


    If $\displaystyle n\geq n_{1}$,then $\displaystyle |t_{n}|<|t|+\epsilon$.................................................. ...........................................3


    If $\displaystyle n\geq n_{2}$,then $\displaystyle |s_{n}-s|<\frac{\epsilon}{2(|t|+\epsilon)}$.................................................. .........................................4


    If $\displaystyle n\geq n_{3}$,then $\displaystyle |t_{n}-t|<\frac{\epsilon}{2(|s|+1)}$.................................................. ...........................................5

    Choose k = max{$\displaystyle n_{1},n_{2},n_{3}$}

    Hence ,

    If $\displaystyle n\geq k$,then $\displaystyle n\geq n_{1},n\geq n_{2},n\geq n_{3}$,

    HENCE

    $\displaystyle |t_{n}|<|t| +\epsilon$.................................................. ........................................6


    $\displaystyle |s_{n}-s|<\frac{\epsilon}{2(|t|+\epsilon)}$.................................................. ........................................7


    $\displaystyle |t_{n}-t|<\frac{\epsilon}{2(|s|+1)}$.................................................. .........................................8

    Now multiply (6) by (7) and we get:


    $\displaystyle |t_{n}||s_{n}-s|< (|t|+\epsilon)\frac{\epsilon}{2(|t|+\epsilon)} = \frac{\epsilon}{2}$.................................................. ...........................................9

    And by multiplying (8) by (|s|+1) we get:


    $\displaystyle (|s|+1)|t_{n}-t|< (|s|+1)\frac{\epsilon}{2(|s|+1)} =\frac{\epsilon}{2}$.................................................. ..........................................10

    Add (9) and (10) we get:


    $\displaystyle |t_{n}||s_{n}-s|+(|s|+1)|t_{n}-t|< \frac{\epsilon}{2} +\frac{\epsilon}{2} = \epsilon$.

    Hence $\displaystyle |t_{n}s_{n}-ts|<\epsilon$.

    Therefor we have proved ,given any +ve No ,ε Τhere exist a natural No k such that :

    If $\displaystyle n\geq k$ then $\displaystyle |t_{n}s_{n}-ts|< \epsilon$ ,for all n
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  6. #6
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    Re the last post.
    Did you note that the OP said not to use the bounded theorem ?
    Your proof uses it.
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  7. #7
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    Nowhere my post uses the the bounded theorem

    Read my post again
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