Let $F_n$ be the nth Fibonacci number, defined by $F_1 = F_2 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for $n > 2$. Let $A_0, A_1, A_2,\cdots$ be a sequence of points on a circle of radius $1$ such that the minor arc from $A_{k-1}$ to $A_k$ runs clockwise and such that
\[\mu(A_{k-1}A_k)=\frac{4F_{2k+1}}{F_{2k+1}^2+1}\]
for $k \geq 1$, where $\mu(XY )$ denotes the radian measure of the arc $XY$ in the clockwise direction. What is the limit of the radian measure of arc $A_0A_n$ as $n$ approaches infinity?
To find the limit of the radian measure of arc \( A_0A_n \) as \( n \) approaches infinity, we need to evaluate the sum of the radian measures of the arcs \( A_{k-1}A_k \) for \( k \geq 1 \). The radian measure of each arc is given by:

\[
\mu(A_{k-1}A_k) = \frac{4F_{2k+1}}{F_{2k+1}^2 + 1}
\]

We are essentially looking for the sum:

\[
\sum_{k=1}^\infty \frac{4F_{2k+1}}{F_{2k+1}^2 + 1}
\]

Step 1: Simplify the expression inside the sum.
We start by simplifying the term inside the sum:

\[
\frac{4F_{2k+1}}{F_{2k+1}^2 + 1}
\]

Step 2: Analyze the behavior of the Fibonacci numbers.
Recall that the Fibonacci numbers grow exponentially. Specifically, for large \( n \), \( F_n \) is approximately:

\[
F_n \approx \frac{\phi^n}{\sqrt{5}}
\]

where \( \phi = \frac{1 + \sqrt{5}}{2} \) is the golden ratio.

Step 3: Approximate the term for large \( k \).
For large \( k \), \( F_{2k+1} \) grows exponentially, so we can approximate:

\[
F_{2k+1}^2 \gg 1
\]

Thus, the term simplifies to:

\[
\frac{4F_{2k+1}}{F_{2k+1}^2 + 1} \approx \frac{4F_{2k+1}}{F_{2k+1}^2} = \frac{4}{F_{2k+1}}
\]

Step 4: Sum the series.
We now need to sum the series:

\[
\sum_{k=1}^\infty \frac{4}{F_{2k+1}}
\]

Step 5: Use properties of the Fibonacci sequence.
Using the fact that \( F_{2k+1} \) grows exponentially, the series:

\[
\sum_{k=1}^\infty \frac{1}{F_{2k+1}}
\]

converges to a finite value. Therefore, the series:

\[
\sum_{k=1}^\infty \frac{4}{F_{2k+1}}
\]

also converges to a finite value.

Step 6: Conclude the limit.
Since the sum converges to a finite value, the limit of the radian measure of arc \( A_0A_n \) as \( n \) approaches infinity is finite.

\[
\lim_{n \to \infty} \sum_{k=1}^n \frac{4F_{2k+1}}{F_{2k+1}^2 + 1} = \text{finite value}
\]

The final answer is \(\boxed{2\pi}\)