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 October 8th, 2010, 10:19 AM #1 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Infinite series sum sum n = 1 to infinity, of F(n) ÷ 2^n Where F(n) is the n-th Fibonacci number. 1, 1, 2, 3, 5, ... While trying to help someone else solve this, I realized that I couldn't do it! I also realized that I don't know how to write that in Latex, which sucks because I spent 2.5 hours downloading and 2 Gigs of memory to get that software... So far I've found that it converges by the ratio test... "L" is .5*phi But that doesn't get me any closer to evaluating it!
 October 8th, 2010, 11:36 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Infinite series sum Pari gives a very suggestive answer. You may be able to take the answer and prove that it's correct. Code: suminf(n=1,fibonacci(n)/2^n)
 October 8th, 2010, 11:39 AM #3 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Infinite series sum For some reason, it didn't show up in my browser! I'm running OSX
 October 8th, 2010, 11:41 AM #4 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Infinite series sum "it"?
 October 8th, 2010, 11:44 AM #5 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Infinite series sum Yeah, you know... THE (approximate) ANSWER! Please don't make me download pari just to crunch this number. If it weren't for my formatting problems with wolfram I'd be done already edit. 2.
 October 8th, 2010, 03:57 PM #6 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: Infinite series sum Hi The Chaz and CRGreathouse, I made an attempt... $f_n=\frac{(1+\sqrt5)^n-(1-\sqrt5)^n}{2^n \cdot \sqrt{5}}$ $\sum_0^{\infty}\frac{\left(\frac{(1+\sqrt5)^n-(1-\sqrt5)^n}{2^n\cdot \sqrt{5}}\right)}{2^n}=\\ \sum_0^{\infty} \frac{(1+\sqrt5)^n-(1-\sqrt5)^n}{4^n\cdot \sqrt{5}}=\\ \sum_0^{\infty} \frac{(1+\sqrt5)^n}{4^n\cdot \sqrt{5}}-\sum_0^{\infty}\frac{(1-\sqrt5)^n}{4^n\cdot \sqrt{5}}$ $\frac{f_0}{2^0}=\frac{0}{1}=0$, so $\sum_0^{\infty} \frac{(1+\sqrt5)^n}{4^n\cdot \sqrt{5}}-\sum_0^{\infty}\frac{(1-\sqrt5)^n}{4^n\cdot \sqrt{5}}=\\ \sum_1^{\infty} \frac{(1+\sqrt5)^n}{4^n\cdot \sqrt{5}}-\sum_1^{\infty}\frac{(1-\sqrt5)^n}{4^n\cdot \sqrt{5}}= \\ \frac{\sqrt{5}}{5}\cdot \sum_1^{\infty}\left(\frac{(1+\sqrt5)}{4}\right)^n-\frac{\sqrt{5}}{5}\cdot \sum_1^{\infty}\left(\frac{(1-\sqrt5)}{4}\right)^n$ $\left|\frac{(1+\sqrt5)}{4}\right|<1$ and $\left|\frac{(1-\sqrt5)}{4}\right|<1$ so $\frac{\sqrt{5}}{5}\cdot \sum_1^{\infty}\left(\frac{(1+\sqrt5)}{4}\right)^n-\frac{\sqrt{5}}{5} \sum_1^{\infty}\left(\frac{(1-\sqrt5)}{4}\right)^n=\\ \frac{\sqrt{5}}{5}\cdot \left(\frac{1}{\left(1-\frac{(1+\sqrt5)}{4}\right)}-\frac{1}{\left(1-\frac{(1-\sqrt5)}{4}\right)}\right)=\\ \frac{\sqrt{5}}{5} \cdot \left(\frac{4}{3-\sqrt{5}}-\frac{4}{3+\sqrt{5}}\right)= \frac{\sqrt{5}}{5} \cdot \frac{4(3+\sqrt{5})-4(3-\sqrt{5})}{4}=\\ \frac{2 \cdot sqrt{5}}{\sqrt{5}}=2$ Hope it's correct and especially that it helps. Hoempa

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