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How can I find the value of the following sum? $$\sum_{n=0}^{\infty}\arctan(\frac{1}{F_{2n+1}})$$ $F_n$ is the Fibonacci number.($F_1=F_2=1$)

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  • $\begingroup$ Just to be sure about the indices, $F_1 = F_2 = 1$? $\endgroup$ – Dan Shved Nov 5 '13 at 9:55
  • $\begingroup$ Well, it clearly is $\frac{\pi}{2}$. Just need to actually prove it ) $\endgroup$ – Dan Shved Nov 5 '13 at 10:03
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OK, let us denote by $a_n$ this complex number: $$a_n = (F_1 + i)(F_3 + i)\ldots(F_{2n-1} + i).$$ I claim that for every $n \geq 1$ we have $a_n = C \cdot (1 + F_{2n} i)$, where $C$ is a positive real number (that depends on $n$).

Let us prove this by induction. For $n = 1$ we have $$a_1 = F_1 + i = 1 + i = 1 + F_2 i.$$

Now the transition. Suppose we have proved that $a_n = C(1 + F_{2n}i)$, where $C$ is a positive real. Then $$ a_{n+1} = a_n (F_{2n+1} + i) = C (1 + F_{2n}i)(F_{2n+1} + i) = C(F_{2n+1}-F_{2n} + i\cdot(F_{2n} F_{2n+1} + 1)). $$ Now, from the equalities on wikipedia it's easy to derive that $F_{2n}F_{2n+1} + 1 = F_{2n-1}F_{2n+2}$. Then we have $$ a_{n+1} = CF_{2n-1}(1 + F_{2n+2}i). $$ $CF_{2n-1}$ is a positive real number, so this completes the proof.

Now we are ready to prove that your infinite sum is equal to $\pi/2$. If we look at the partial sum, we easily find that $$ \sum_{n=0}^{k}\arctan(\frac{1}{F_{2n+1}}) = \sum_{n=0}^{k}\arg (F_{2n+1} + i) = \arg a_{k+1} = \arctan(F_{2k + 2}). $$ As $k$ tends to $+\infty$, $F_{2k+2}$ also tends to $+\infty$, and its $\arctan$ tends to $\pi/2$. So the answer is $\pi/2$.

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  • $\begingroup$ +1 Nice use of complex numbers! I was so sold on sticking to Fibonacci identities, and the basic identities of tangent. $\endgroup$ – Jyrki Lahtonen Nov 5 '13 at 10:51
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We can prove by induction that $$ \sum_{n=0}^k\arctan\frac1{F_{2n+1}}=\arctan F_{2k+2}. $$ The base case $k=0$ is immediate, because $F_1=F_2=1$.

OTOH by induction hypothesis $$ \sum_{n=0}^{k+1}\arctan\frac1{F_{2n+1}}=\arctan F_{2k+2}+\arctan\frac1{F_{2k+3}}. $$ It follows from the formula for the tangent of the sum of two angles (careful about the overflow!) $$ \arctan x+\arctan y\equiv \arctan\frac{x+y}{1-xy}\pmod\pi. $$ Here $x=F_{2k+2}$, $y=1/F_{2k+3}$, and thus $$ \begin{aligned} \frac{x+y}{1-xy}&=\frac{F_{2k+2}F_{2k+3}+1}{F_{2k+3}-F_{2k+2}}\\ &=\frac{F_{2k+2}F_{2k+3}+1}{F_{2k+1}}=F_{2k+4} \end{aligned} $$ by the identity $F_{2k+4}F_{2k+1}=1+F_{2k+2}F_{2k+3}$ completing our induction step (see the link given by Dan Shved or prove this identity by induction).

As $F_n\to\infty$ it follows that the limit is $\pi/2$.

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  • $\begingroup$ A cute problem. Thanks to whoever came up with this. I want to use this as an extra HW problem some day. $\endgroup$ – Jyrki Lahtonen Nov 5 '13 at 10:52
  • $\begingroup$ I kind of cheated in my answer by not mentioning the possibility of an overflow. Thanks for bringing that up, I guess :) $\endgroup$ – Dan Shved Nov 5 '13 at 10:59
  • $\begingroup$ Well, I didn't really deal with it either. Adding one $\arctan$ at a time, as in an induction, sorta prevents the overflows I think :-/ $\endgroup$ – Jyrki Lahtonen Nov 5 '13 at 11:02

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