A series involving Fibanacci

Let $F_n$ denote the $n$-th Fibonacci number with initial values $F_1=F_2=1$. Prove that:

$$\sum_{n=0}^{\infty} \arctan \frac{1}{F_{2n+1}} = \frac{\pi}{2}$$

Solution

I kinda recalled the exercise when today I saw this question posted at mathematica.gr by Demetres. The series actually telescoped but here is a bit more difficult to actually manipulate our sum. Anyway our solution will invoke Cassini - Catalan identity. So let us get the show on the road shall we?

Our primary goal is to actually re-write the summand in the form:

$$\frac{a_{n+1}-a_n}{1+a_n a_{n+1}} = \frac{1}{F_{2n+1}}$$

Well using the Cassini - Catalan identity which states that $\displaystyle F_{2n+1}^2 =1+F_{2n+2}F_{2n}$ then we can easily see that the $a_n$ sequence that serves our needs is nothing else than $a_n=F_{2n}$. Thus:

\begin{equation} \frac{a_{n+1}-a_n}{1+a_n a_{n+1}} = \frac{F_{2n+2}-F_{2n}}{1+F_{2n+2}F_{2n}} = \frac{F_{2n+1}}{F_{2n+1}^2} = \frac{1}{F_{2n+1}} \end{equation}

So using $(1)$ our sum can be evaluated as follows:

\begin{align*}
\sum_{n=0}^{\infty} \arctan \frac{1}{F_{2n+1}} &=\sum_{n=0}^{\infty} \left [ \arctan F_{2n+2} - \arctan F_{2n} \right ] \\
 &= \lim_{m \rightarrow +\infty} \sum_{n=0}^{m} \left [ \arctan F_{2n+2} - \arctan F_{2n} \right ]  \\
 &= \lim_{m \rightarrow +\infty}\arctan F_{2m+2}\\
 &= \frac{\pi}{2}
\end{align*}

The reader is left to verify that $\displaystyle \sum_{n=0}^{m} \left [ \arctan F_{2n+2} - \arctan F_{2n} \right ] = \arctan F_{2m+2}$ which is quite easy because it telescopes.