T thatguy47 Junior Member Joined Aug 11, 2008 Messages 69 Jan 21, 2009 #1 Determine whether the sequence converges or diverges. if it converges find the limit. How would you go about solving this? It's different than the rest of the problems I've been doing and there's no examples like this in my notes.
Determine whether the sequence converges or diverges. if it converges find the limit. How would you go about solving this? It's different than the rest of the problems I've been doing and there's no examples like this in my notes.
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Jan 21, 2009 #2 Rewrite it as: \(\displaystyle \frac{1}{3}\cdot a_{n}\) \(\displaystyle \frac{1}{3}\cdot \frac{2^{n}}{3^{n}}\) \(\displaystyle \frac{1}{3}\cdot (\frac{2}{3})^{n}\) Now, can you see what it does?. See the limit?. List out some terms if it helps. If the even numbered terms and the odd numbered terms both converge to a limit than so does the sequence.
Rewrite it as: \(\displaystyle \frac{1}{3}\cdot a_{n}\) \(\displaystyle \frac{1}{3}\cdot \frac{2^{n}}{3^{n}}\) \(\displaystyle \frac{1}{3}\cdot (\frac{2}{3})^{n}\) Now, can you see what it does?. See the limit?. List out some terms if it helps. If the even numbered terms and the odd numbered terms both converge to a limit than so does the sequence.
T thatguy47 Junior Member Joined Aug 11, 2008 Messages 69 Jan 21, 2009 #3 Could you just explain where the 1/3 comes from?
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Jan 21, 2009 #4 Uh oh. You don't see that?. Doesn't \(\displaystyle 3^{n+1}=3\cdot 3^{n}\)?.
T thatguy47 Junior Member Joined Aug 11, 2008 Messages 69 Jan 21, 2009 #5 Ya...I didn't see that, I completely forgot learning that. Thanks.