# Math Help - Sequences and their Limits

1. ## Sequences and their Limits

Use Theorem 3 to show that the sequence 0.7, 0.77, 0.777, 0.7777, 0.77777, 0.777777, 0.7777777, ... has a limit.

Theorem 3 is ...

Suppose that the sequence $a_{k}$ is nondecresing and bounded above by a number A. That is,

$a_{1} \leq a_{2} \leq a_{3} \leq A$

Then $a_{k}$ converges to some finite limit a, with a $\leq A$. Similarly, if $b_{k}$ is nonincreasing and bounded below by a number B, then $b_{k}$ converges to a finite limit $b \geq B$.

edit: How do I do the less than or equal to sign in LaTeX?

2. Originally Posted by larson
Use Theorem 3 to show that the sequence 0.7, 0.77, 0.777, 0.7777, 0.77777, 0.777777, 0.7777777, ... has a limit.

Theorem 3 is ...

Suppose that the sequence $a_{k}$ is nondecresing and bounded above by a number A. That is,

$a_{1}$ <, $a_{2}$ < $a_{3}$ < A

Then $a_{k}$ converges to some finite limit a, with a < A. Similarly, if $b_{k}$ is nonincreasing and bounded below by a number B, then $b_{k}$ converges to a finite limit b > B.

edit: How do I do the less than or equal to sign in LaTeX?
$/leq$ gives $\leq$ just change / to \

3. Originally Posted by larson
Use Theorem 3 to show that the sequence 0.7, 0.77, 0.777, 0.7777, 0.77777, 0.777777, 0.7777777, ... has a limit.

Theorem 3 is ...

Suppose that the sequence $a_{k}$ is nondecresing and bounded above by a number A. That is,

$a_{1}$ <, $a_{2}$ < $a_{3}$ < A

Then $a_{k}$ converges to some finite limit a, with a < A. Similarly, if $b_{k}$ is nonincreasing and bounded below by a number B, then $b_{k}$ converges to a finite limit b > B.

edit: How do I do the less than or equal to sign in LaTeX?
you can use the first part of the theorem with $A = 0. \bar{7}$ or $A = 0.8$

4. Originally Posted by larson
Use Theorem 3 to show that the sequence 0.7, 0.77, 0.777, 0.7777, 0.77777, 0.777777, 0.7777777, ... has a limit.

Theorem 3 is ...

Suppose that the sequence $a_{k}$ is nondecresing and bounded above by a number A. That is,

$a_{1}$ <, $a_{2}$ < $a_{3}$ < A

Then $a_{k}$ converges to some finite limit a, with a < A. Similarly, if $b_{k}$ is nonincreasing and bounded below by a number B, then $b_{k}$ converges to a finite limit b > B.

edit: How do I do the less than or equal to sign in LaTeX?
and I will give you a hint $a_n=7\sum_{t=1}^{n}\cdot\bigg(\frac{1}{10}\bigg)^{ t}$

5. Originally Posted by Mathstud28
$/leq$ gives $\leq$ just change / to \
it suffices to type le, likewise, ge and ne for greater thank and not equal to respectively