# Help on a convergent sequence limit problem

• Jan 6th 2010, 07:40 PM
Kitty216
Help on a convergent sequence limit problem
http://i146.photobucket.com/albums/r...ty216/prob.jpg

I tried using the method of doing a limit to infinity which gave me infinity/infinity (an indeterminate) so I used L'hopital's rule and took the derivative of top and bottom. However that still leaves me with that segment. Some help would be nice, thanks.
• Jan 6th 2010, 07:43 PM
Drexel28
Quote:

Originally Posted by Kitty216
http://i146.photobucket.com/albums/r...ty216/prob.jpg

I tried using the method of doing a limit to infinity which gave me infinity/infinity (an indeterminate) so I used L'hopital's rule and took the derivative of top and bottom. However that still leaves me with that segment. Some help would be nice, thanks.

Really? You know L'hopital's rule but you don't know that $|k|<1\implies\lim k^n=0$?
• Jan 6th 2010, 07:44 PM
Kitty216
i do not know that sorry =/
• Jan 6th 2010, 07:50 PM
Drexel28
Quote:

Originally Posted by Kitty216
i do not know that sorry =/

There's no need to be sorry! It makes sense though, right? If you take an number less than one, then squaring it makes it smaller. Cubing it makes it even smaller, ad infinitum.
• Jan 6th 2010, 07:52 PM
ANDS!
Yes you do, though you might not immediately recognize it. What he is saying is that if the absolute value of "k" is less than 1, then the limit as n approaches infinity of k to the n, is 0.

$a_{n}=\frac{3^n}{4^n} \Longrightarrow a_{n}=\left(\frac{3}{4}\right)^{n}$

Where k is equal to $\frac{3}{4}$
• Jan 6th 2010, 07:53 PM
Kitty216
oh yea that does make sense since the bottom would keep getting smaller than the top. However my teacher says I have to do it in an algebraic way and show work. Is there really a way with work to show?
• Jan 6th 2010, 07:57 PM
Drexel28
Quote:

Originally Posted by Kitty216
oh yea that does make sense since the bottom would keep getting smaller than the top. However my teacher says I have to do it in an algebraic way and show work. Is there really a way with work to show?

I mean you can prove it by the definition. Or note that $\lim_{n\to\infty}\frac{\left(\frac{3}{4}\right)^{n +1}}{\left(\frac{3}{4}\right)^n}=\frac{3}{4}<1$ so that $\sum_{n=1}^{\infty}\left(\frac{3}{4}\right)^n$ converges which means that $\left(\frac{3}{4}\right)^n\to 0$
• Jan 6th 2010, 08:04 PM
Kitty216
hmm I see... alrighty. Thanks for the help! (Wink)