# Infinite Series $\sum\limits_{n=0}^{\infty}\arctan(\frac{1}{F_{2n+1}})$

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.($F_1=F_2=1$)

• Just to be sure about the indices, $F_1 = F_2 = 1$? – Dan Shved Nov 5 '13 at 9:55
• Well, it clearly is $\frac{\pi}{2}$. Just need to actually prove it ) – Dan Shved Nov 5 '13 at 10:03

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 $\pi/2$.

• +1 Nice use of complex numbers! I was so sold on sticking to Fibonacci identities, and the basic identities of tangent. – Jyrki Lahtonen Nov 5 '13 at 10:51

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 $\pi/2$.

• A cute problem. Thanks to whoever came up with this. I want to use this as an extra HW problem some day. – Jyrki Lahtonen Nov 5 '13 at 10:52
• I kind of cheated in my answer by not mentioning the possibility of an overflow. Thanks for bringing that up, I guess :) – Dan Shved Nov 5 '13 at 10:59
• 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 :-/ – Jyrki Lahtonen Nov 5 '13 at 11:02