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1) Let $F_1 = 1$, $F_2 = 1$, and $F_n = F_{n-1} + F_{n-2}$ for $n \geq 3$.

How do you prove that

$$\sum_{n=2}^\infty \frac1{F_{n-1} F_{n+1}} = 1$$

$$\sum_{n=2}^\infty \frac{F_n}{F_{n-1} F_{n+1}} = 1$$

Do you even use the formula $n(n+1)/2$?

I tried using $F_{n-1} = F_n - F_{n-2}$ and $F_{n+1} = F_n + F_{n-1}$ and then trying to figure out how to substitute those, but I've been going around in circles. How would you solve this? What tips can you give me to solve this? Is there any website that teaches this? I tried searching google and youtube but none were helpful.

2) Let $s_x = \tan(2/x)$. Where does this become nonincreasing? While using monotonic limits to establish there is a limit.

I was guessing you need to use upperbound and lowerbound but doesn't it go to infinity?

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The second Fibonacci sum should be 2, not 1. – John Bentin Jan 18 '13 at 1:26
up vote 6 down vote accepted

For the first one, note that $F_n = F_{n+1} - F_{n-1}$. Hence, $$\dfrac1{F_{n-1} F_{n+1}} = \dfrac1{F_n F_{n-1}} - \dfrac1{F_n F_{n+1}}$$ and similarly $$\dfrac{F_n}{F_{n+1} F_{n-1}} = \dfrac1{F_{n-1}} - \dfrac1{F_{n+1}}$$ Now use telescopic summation and cancel terms to get what you want.

For the second one, note that $\tan(x)$ is increasing in the region $[0,\pi/2)$. Hence, $\tan(2/x)$ is decreasing for $x>0$. Hence, you sequence $s_n = \tan(2/n)$ is a monotone decreasing sequence and the limit is $0$ since $\tan(x)$ is continuous near origin and $\tan(0) = 0$.

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for telescopic summation. are you suggesting i find the values of all those functions and then some how find a value n so that i can use limit laws? – user1730308 Jan 18 '13 at 3:17
@user1730308 Write down the first few terms and add them and see what happens. Also note that $F_n \to \infty$ as $n \to \infty$ i.e. $\dfrac1{F_n} \to \infty$. – user17762 Jan 18 '13 at 3:18
one last question shouldnt question number to be x<0 because tan becomes decreasing if y=-tanx or y=tan(-x)? thanks – user1730308 Jan 18 '13 at 16:25

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