# Thread: Sequence Convergence Using Definition of Convergence

1. ## 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....

2. 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 $\frac{2}{\varepsilon } < N$.

3. Originally Posted by Plato
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?

4. $\left| {\frac{{2 - 2n}}
{n} + 2} \right| = \left| {\frac{2}
{n}} \right| = \left| 2 \right|\left| {\frac{1}
{n}} \right| < 2\frac{\varepsilon }
{2} = \varepsilon$

5. Originally Posted by Plato
$\left| {\frac{{2 - 2n}}
{n} + 2} \right| = \left| {\frac{2}
{n}} \right| = \left| 2 \right|\left| {\frac{1}
{n}} \right| < 2\frac{\varepsilon }
{2} = \varepsilon$
How does this show that it converges to -2?

6. Originally Posted by kathrynmath
How does this show that it converges to -2?
$\left| {\frac{{2 - 2n}}
{n} + 2} \right| = \left| {\left( {\frac{{2 - 2n}}
{n}} \right) - \left( { - 2} \right)} \right|$

Are you sure that you really understand the definition?

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

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