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Math Help - Sequence Convergence Using Definition of Convergence

  1. #1
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    Sequence Convergence Using Definition of Convergence

    {(2-2n)/n}
    I want to use the definition of convergence to prove this converges. So I assume the sequence converges to -2. Let epsilon>0 be arbitrary. We need to find N so that when ever n>N then absolute value of(a subn-A)=
    abs value((2-2n)/n+2)<epsilon
    -epsilon<2/n-2n/n+2<epsilon
    -epsilon<2/n<epsilon
    -2/epsilon>n>2/epsilon=N

    This is where I get stuck....
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  2. #2
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    Quote Originally Posted by kathrynmath View Post
    {(2-2n)/n}
    I want to use the definition of convergence to prove this converges. So I assume the sequence converges to -2. Let epsilon>0 be arbitrary. We need to find N so that when ever n>N then absolute value of(a subn-A)=
    abs value((2-2n)/n+2)<epsilon
    -epsilon<2/n-2n/n+2<epsilon
    -epsilon<2/n<epsilon
    -2/epsilon>n>2/epsilon=N
    Find an N such that \frac{2}{\varepsilon } < N .
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  3. #3
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    Quote Originally Posted by Plato View Post
    Find an N such that \frac{2}{\varepsilon } < N .
    so, 1/n<1/N<epsilon
    1/n<1/(2/epsilon)<epsilon
    1/n<epsilon/2<epsilon

    Is this what I do?
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  4. #4
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    \left| {\frac{{2 - 2n}}<br />
{n} + 2} \right| = \left| {\frac{2}<br />
{n}} \right| = \left| 2 \right|\left| {\frac{1}<br />
{n}} \right| < 2\frac{\varepsilon }<br />
{2} = \varepsilon
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  5. #5
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    Quote Originally Posted by Plato View Post
    \left| {\frac{{2 - 2n}}<br />
{n} + 2} \right| = \left| {\frac{2}<br />
{n}} \right| = \left| 2 \right|\left| {\frac{1}<br />
{n}} \right| < 2\frac{\varepsilon }<br />
{2} = \varepsilon
    How does this show that it converges to -2?
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  6. #6
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    Quote Originally Posted by kathrynmath View Post
    How does this show that it converges to -2?
    \left| {\frac{{2 - 2n}}<br />
{n} + 2} \right| = \left| {\left( {\frac{{2 - 2n}}<br />
{n}} \right) - \left( { - 2} \right)} \right|

    Are you sure that you really understand the definition?
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  7. #7
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    Quote Originally Posted by Plato View Post
    \left| {\frac{{2 - 2n}}<br />
{n} + 2} \right| = \left| {\left( {\frac{{2 - 2n}}<br />
{n}} \right) - \left( { - 2} \right)} \right|

    Are you sure that you really understand the definition?
    never mind, I see what you meant...
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