# Sequence Convergence Using Definition of Convergence

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• Oct 24th 2008, 11:49 AM
kathrynmath
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....
• Oct 24th 2008, 12:35 PM
Plato
Quote:

Originally Posted by kathrynmath
{(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 $\displaystyle \frac{2}{\varepsilon } < N$.
• Oct 24th 2008, 12:44 PM
kathrynmath
Quote:

Originally Posted by Plato
Find an N such that $\displaystyle \frac{2}{\varepsilon } < N$.

so, 1/n<1/N<epsilon
1/n<1/(2/epsilon)<epsilon
1/n<epsilon/2<epsilon

Is this what I do?
• Oct 24th 2008, 12:58 PM
Plato
$\displaystyle \left| {\frac{{2 - 2n}} {n} + 2} \right| = \left| {\frac{2} {n}} \right| = \left| 2 \right|\left| {\frac{1} {n}} \right| < 2\frac{\varepsilon } {2} = \varepsilon$
• Oct 24th 2008, 01:19 PM
kathrynmath
Quote:

Originally Posted by Plato
$\displaystyle \left| {\frac{{2 - 2n}} {n} + 2} \right| = \left| {\frac{2} {n}} \right| = \left| 2 \right|\left| {\frac{1} {n}} \right| < 2\frac{\varepsilon } {2} = \varepsilon$

How does this show that it converges to -2?
• Oct 24th 2008, 01:25 PM
Plato
Quote:

Originally Posted by kathrynmath
How does this show that it converges to -2?

$\displaystyle \left| {\frac{{2 - 2n}} {n} + 2} \right| = \left| {\left( {\frac{{2 - 2n}} {n}} \right) - \left( { - 2} \right)} \right|$

Are you sure that you really understand the definition?
• Oct 24th 2008, 01:45 PM
kathrynmath
Quote:

Originally Posted by Plato
$\displaystyle \left| {\frac{{2 - 2n}} {n} + 2} \right| = \left| {\left( {\frac{{2 - 2n}} {n}} \right) - \left( { - 2} \right)} \right|$

Are you sure that you really understand the definition?

never mind, I see what you meant...