# Thread: series and sequence help

1. ## series and sequence help

sorry for the image but i am having trouble proving the answers i get while doing my first problem

(a) I know the sequence converges to 0 since if you take the limit as n goes to infinity, then a(n) will go to 0 because fractions multiplied by themselves many times that are below 1 go to 0. but how do i prove this answer??

(b) is this geometric?

(c) does it converge because its a constant ratio by geometric series test?

(d) does it converge to 9?

(e) i know that the answer is yes, .999... is equal to 1. but how do i prove that? would i set up a series of some kind??

thanks

2. Originally Posted by Arez

(b) is this geometric?

For your series $\{t_1,t_2,t_3,\cdots\} = \{\left(\frac{9}{10}\right)^1,\left(\frac{9}{10}\r ight)^2,\left(\frac{9}{10}\right)^3,\cdots\}$

It is geometric if $\frac{t_3}{t_2} = \frac{t_2}{t_1}$

3. Originally Posted by Arez

(c) does it converge because its a constant ratio by geometric series test?

Only if

$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right |<1$

4. (e) Let
$x=.\bar 9$
then
$10x=9.\bar 9$
subtract x from both sides, x from the left and $.\bar 9$ from the right

so $9x=9$ or x=1.

5. (b) geometric with r=.9, Geo's converge when -1<r<1.
The ratio test is basically a limit comparison to a geometric.
The limit of adjacent terms is the search for the essential r.

(c) and (d) yes it converges to 9...

the limit is

${9/10\over 1-9/10}={9\over 10-9}=9$

6. (a) $(.9)^n\to 0$ as n goes to infinity

7. Yes, the original poster said he knew that in his original post.