<Question>
Compute
\[\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}},\]where $F_n$ denotes the $n$th Fibonacci number, so $F_0 = 0$ and $F_1 = 1.$
</Question>

<think>
<Plan_1>
Okay, so I need to compute the infinite sum \(\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}}\), where \(F_n\) are the Fibonacci numbers with \(F_0 = 0\) and \(F_1 = 1\). Hmm, let's start by recalling what the Fibonacci sequence looks like. The Fibonacci numbers are defined by \(F_0 = 0\), \(F_1 = 1\), and each subsequent term is the sum of the two previous ones: \(F_{n} = F_{n-1} + F_{n-2}\) for \(n \geq 2\). So, the sequence goes 0, 1, 1, 2, 3, 5, 8, 13, 21, and so on.

First, let me write out the first few terms of the series to see if I can spot a pattern or a telescoping nature, which might be helpful for summing the series.

<prompt_1.1> Write out the first few terms of the series \(\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}}\) to identify any patterns or telescoping behavior. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
The general term of the series is \(\frac{F_{n + 1}}{F_n F_{n + 2}}\). Let's compute this for the first few values of \(n\):

For \(n = 1\):
Numerator: \(F_{2} = 1\)
Denominator: \(F_1 F_3 = 1 \times 2 = 2\)
So the term is \(1/2\).

For \(n = 2\):
Numerator: \(F_{3} = 2\)
Denominator: \(F_2 F_4 = 1 \times 3 = 3\)
Term: \(2/3\)

For \(n = 3\):
Numerator: \(F_4 = 3\)
Denominator: \(F_3 F_5 = 2 \times 5 = 10\)
Term: \(3/10\)

For \(n = 4\):
Numerator: \(F_5 = 5\)
Denominator: \(F_4 F_6 = 3 \times 8 = 24\)
Term: \(5/24\)

For \(n = 5\):
Numerator: \(F_6 = 8\)
Denominator: \(F_5 F_7 = 5 \times 13 = 65\)
Term: \(8/65\)

So the first few terms of the series are \(1/2 + 2/3 + 3/10 + 5/24 + 8/65 + \dots\)
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Hmm, I wonder if this series telescopes. Telescoping series often have terms that cancel out when written in a certain way. To check this, maybe I need to manipulate the general term \(\frac{F_{n + 1}}{F_n F_{n + 2}}\) to express it as a difference of two fractions, which might telescope when summed.

Let me think about identities involving Fibonacci numbers. One famous identity is Cassini's identity: \(F_{n+1}F_{n-1} - F_n^2 = (-1)^n\). Not sure if that's directly useful here, but maybe there's another identity that could help.

Alternatively, perhaps using the ratio of consecutive Fibonacci numbers. Since Fibonacci numbers grow exponentially (approximately like \(\phi^n\), where \(\phi = \frac{1+\sqrt{5}}{2}\)), the terms might be decreasing rapidly enough for the series to converge, but since it's an infinite sum, we need to find a closed-form expression.

Let me consider writing the term \(\frac{F_{n + 1}}{F_n F_{n + 2}}\) in terms of \(\frac{1}{F_n}\) and \(\frac{1}{F_{n + 2}}\).

<prompt_2.1> Consider and test the approach of expressing the term \(\frac{F_{n + 1}}{F_n F_{n + 2}}\) as a difference of reciprocals of Fibonacci numbers. </prompt_2.1>

Let me compute this for the first few terms to see if it aligns with the original series.

<prompt_2.2> Compute the partial sums of the series using the expression \(1/F_n - 1/F_{n+1}\) and compare them to the partial sums of the original series. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Wait, let's see:

We have \(F_{n + 2} = F_{n + 1} + F_n\). So, maybe we can express the denominator in terms of \(F_{n + 1}\) and \(F_n\).

Alternatively, perhaps using the identity \(F_{n + 2} = F_{n + 1} + F_n\), so \(1/F_{n + 2} = 1/(F_{n + 1} + F_n)\). Not sure if that helps. Alternatively, maybe consider the reciprocal of the denominator:

\(\frac{1}{F_n F_{n + 2}} = \frac{1}{F_n (F_{n + 1} + F_n)}\). Hmm, perhaps partial fractions? Let me try to decompose this fraction.

Suppose we can write \(\frac{1}{F_n (F_{n + 1} + F_n)}\) as \(A \left( \frac{1}{F_n} - \frac{1}{F_{n + 1}} \right)\) or something similar. Let's test this idea. Let me assume that:

\[
\frac{1}{F_n (F_{n + 1} + F_n)} = A \left( \frac{1}{F_n} - \frac{1}{F_{n + 1}} \right)
\]

Let's solve for \(A\):

Multiply both sides by \(F_n F_{n + 1}\):

Left side: \(F_{n + 1}\)

Right side: \(A (F_{n + 1} - F_n)\)

Therefore:

\(F_{n + 1} = A (F_{n + 1} - F_n)\)

But \(F_{n + 1} - F_n = F_{n - 1}\), since \(F_{n + 1} = F_n + F_{n - 1}\). Therefore:

\(F_{n + 1} = A F_{n - 1}\)

Hence, \(A = \frac{F_{n + 1}}{F_{n - 1}}\). Hmm, but \(A\) is supposed to be a constant, not depending on \(n\). So unless \(\frac{F_{n + 1}}{F_{n - 1}}\) is a constant, which it isn't (since Fibonacci numbers grow exponentially), this approach might not work. So maybe partial fractions in this way isn't the right approach.
</execution_2.1>

<execution_2.2>
Sum up to n=1: 1/2. According to the partial sums of the transformed series: 1/F_1 - 1/F_3 = 1/1 - 1/2 = 1 - 0.5 = 0.5. That matches.

Sum up to n=2: 1/2 + 2/3 = 5/6 ≈ 0.8333. The partial sum of the transformed series: 1/F_1 - 1/F_4 = 1/1 - 1/3 ≈ 0.6667. Wait, that's 2/3, but the actual sum up to n=2 is 1/2 + 2/3 = 5/6. Hmm, discrepancy here. So maybe my initial idea is missing something.

Wait, perhaps the partial sums of the transformed series are 1/F_n - 1/F_{n+1}, so when summed from n=1 to N, the total would be (1/F_1 - 1/F_2) + (1/F_2 - 1/F_3) + ... + (1/F_N - 1/F_{N+1}) ) = 1/F_1 - 1/F_{N+1}. Since F_1 = 1, this would be 1 - 1/F_{N+1}. But if that's the case, then the infinite sum would be the limit as N approaches infinity of 1 - 1/F_{N+1} = 1 - 0 = 1. But wait, when I computed the partial sums up to n=2, the actual sum is 5/6, but according to this telescoping series, the partial sum should be 1 - 1/F_3 = 1 - 1/2 = 1/2, which doesn't match. So there must be a mistake here.

Wait, perhaps the decomposition is different. Let me check again.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1, execution_2.2:
Wait, if I sum from n=1 to N the term \(1/F_n - 1/F_{n+1}\), then the sum telescopes to \(1/F_1 - 1/F_{N+1}\). Since \(F_1 = 1\), this is \(1 - 1/F_{N+1}\). Therefore, as \(N \to \infty\), \(F_{N+1} \to \infty\), so the sum converges to \(1 - 0 = 1\).

But when I computed the partial sums for the original series:

<prompt_3.1> Re-evaluate the partial sums for the original series and identify any miscalculations. </prompt_3.1>

Wait, that's a problem. There's a discrepancy here. If the transformed series telescopes to 1 - 1/F_{N+1}, but the original series' partial sums are different. So perhaps my initial approach is wrong.

Wait, so let's re-examine. The term in the series is \(\frac{F_{n+1}}{F_n F_{n+2}}\). If I can express this as \( \frac{1}{F_n} - \frac{1}{F_{n+2}} \), then summing over n would telescope. Let me check if that's possible.

Suppose:

<prompt_3.2> Re-examine the approach to express the term \(\frac{F_{n+1}}{F_n F_{n+2}}\) as a telescoping difference and verify with the first few terms. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
n=1: 1/2 ≈ 0.5

n=1-2: 1/2 + 2/3 ≈ 0.5 + 0.6667 ≈ 1.1667, but according to the telescoping sum, it should be 1 - 1/3 ≈ 0.6667. That's a conflict. Therefore, my assumption that \(\frac{F_{n+1}}{F_n F_{n+2}} = \frac{1}{F_n} - \frac{1}{F_{n+1}}\) is incorrect.
</execution_3.1>

<execution_3.2>
\[
\frac{1}{F_n} - \frac{1}{F_{n+2}} = \frac{F_{n+2} - F_n}{F_n F_{n+2}}}
\]

But \(F_{n+2} - F_n = F_{n+1}\), since \(F_{n+2} = F_{n+1} + F_n\), so \(F_{n+2} - F_n = F_{n+1}\). Therefore:

\[
\frac{1}{F_n} - \frac{1}{F_{n+2}} = \frac{F_{n+1}}{F_n F_{n+2}}}
\]

Wait, that's exactly the term in our series! So that means:

\[
\frac{F_{n+1}}{F_n F_{n+2}} = \frac{1}{F_n} - \frac{1}{F_{n+2}}
\]

Wow, that's a key insight. So each term in the series can be written as the difference of reciprocals of Fibonacci numbers two indices apart. Therefore, the series becomes:

\[
\sum_{n = 1}^\infty \left( \frac{1}{F_n} - \frac{1}{F_{n+2}} \right )
\]

Now, let's write out the first few terms of this telescoping series:

For \(n = 1\): \(\frac{1}{F_1} - \frac{1}{F_3} = 1 - \frac{1}{2}\)

For \(n = 2\): \(\frac{1}{F_2} - \frac{1}{F_4} = \frac{1}{1} - \frac{1}{3}\)

For \(n = 3\): \(\frac{1}{F_3} - \frac{1}{F_5} = \frac{1}{2} - \frac{1}{5}\)

For \(n = 4\): \(\frac{1}{F_4} - \frac{1}{F_6} = \frac{1}{3} - \frac{1}{8}\)

For \(n = 5\): \(\frac{1}{F_5} - \frac{1}{F_7} = \frac{1}{5} - \frac{1}{13}\)

So when we add these up, let's see the partial sum up to \(n = 5\):

\[
\left(1 - \frac{1}{2}\right) + \left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{5}\right) + \left(\frac{1}{3} - \frac{1}{8}\right) + \left(\frac{1}{5} - \frac{1}{13}\right)
\]

Let's rearrange the terms:

\(1 + 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{5} - \frac{1}{2} - \frac{1}{3} - \frac{1}{8} - \frac{1}{13}\)

Wait, no, more carefully:

First term: 1 - 1/2

Second term: 1 - 1/3

Third term: 1/2 - 1/5

Fourth term: 1/3 - 1/8

Fifth term: 1/5 - 1/13

Adding them up:

1 - 1/2 + 1 - 1/3 + 1/2 - 1/5 + 1/3 - 1/8 + 1/5 - 1/13

Now, let's cancel terms:

- The -1/2 cancels with +1/2

- The -1/3 cancels with +1/3

- The -1/5 cancels with +1/5

So we are left with:

1 + 1 - 1/8 - 1/13

Which is 2 - 1/8 - 1/13

But wait, that's for n=5. Wait, but actually, when summing up to n=5, the partial sum would be:

Sum = (1 - 1/2) + (1 - 1/3) + (1/2 - 1/5) + (1/3 - 1/8) + (1/5 - 1/13)

Grouping all the terms:

1 + 1 + ( -1/2 + 1/2 ) + ( -1/3 + 1/3 ) + ( -1/5 + 1/5 ) - 1/8 - 1/13

So indeed, all the middle terms cancel, leaving:

1 + 1 - 1/8 - 1/13 = 2 - (1/8 + 1/13)

But according to the telescoping series, the partial sum up to N should be:

Sum from n=1 to N of (1/F_n - 1/F_{n+2}) = (1/F_1 + 1/F_2) - (1/F_{N+1} + 1/F_{N+2})

Wait, let's check that. Let's consider the general case.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.2:
The sum from n=1 to N of (1/F_n - 1/F_{n+2}) can be written as:

<prompt_4.1> Derive the general expression for the partial sum of the telescoping series and compute the limit as \(N\) approaches infinity. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Sum = (1/F_1 - 1/F_3) + (1/F_2 - 1/F_4) + (1/F_3 - 1/F_5) + ... + (1/F_N - 1/F_{N+2})

When we add these up, most terms cancel:

- The -1/F_3 cancels with +1/F_3

- Similarly, -1/F_4 cancels with +1/F_4 if it appears, but in the next term, we have -1/F_4 and +1/F_4 from the term after. Wait, let's see:

Wait, let's list the terms:

n=1: 1/F_1 - 1/F_3

n=2: 1/F_2 - 1/F_4

n=3: 1/F_3 - 1/F_5

n=4: 1/F_4 - 1/F_6

... 

n=N: 1/F_N - 1/F_{N+2}

So when we sum from n=1 to N, the total sum is:

1/F_1 + 1/F_2 + ( -1/F_3 + 1/F_3 ) + ( -1/F_4 + 1/F_4 ) + ... + ( -1/F_{N+1} + 1/F_{N+1} ) + ( -1/F_{N+2} )

Wait, but in the terms above, after n=1 and n=2, the next terms cancel as we go along. Wait, actually, the cancellation is:

- The -1/F_3 from n=1 cancels with the +1/F_3 from n=3

- The -1/F_4 from n=2 cancels with the +1/F_4 from n=4

- This pattern continues up to n=N-1 and n=N.

Wait, but for n=1 to N, the terms would cancel as follows:

Sum = 1/F_1 + 1/F_2 + ( -1/F_3 + 1/F_3 ) + ( -1/F_4 + 1/F_4 ) + ... + ( -1/F_{N+1} + 1/F_{N+1} ) + ( -1/F_{N+2} )

Wait, but actually, for each k from 3 to N+1, the term -1/F_k cancels with a +1/F_k from a different term. However, since the upper limit is N, the last term is 1/F_N - 1/F_{N+2}. Therefore, the total sum would be:

Sum = 1/F_1 + 1/F_2 - 1/F_{N+1} - 1/F_{N+2}

Because all the intermediate terms from 1/F_3 up to 1/F_{N} get canceled in pairs, except for the first two terms (1/F_1 and 1/F_2) and the last two negative terms (-1/F_{N+1} and -1/F_{N+2}).

Given that F_1 = 1 and F_2 = 1, this simplifies to:

Sum = 1 + 1 - 1/F_{N+1} - 1/F_{N+2} = 2 - (1/F_{N+1} + 1/F_{N+2})

Therefore, as N approaches infinity, the terms 1/F_{N+1} and 1/F_{N+2} approach zero because the Fibonacci numbers grow exponentially. Hence, the sum converges to 2 - 0 = 2.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
But wait, let's verify this with the partial sums we computed earlier.

<prompt_5.1> Verify the derived formula for the partial sum with the computed partial sums and confirm the limit as \(N\) approaches infinity. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
For N=1:

Sum = 1 + 1 - 1/F_2 - 1/F_3 = 2 - 1/1 - 1/2 = 2 - 1 - 0.5 = 0.5, which matches the first term.

For N=2:

Sum = 2 - 1/F_3 - 1/F_4 = 2 - 1/2 - 1/3 ≈ 2 - 0.5 - 0.333... ≈ 1.1666...., which matches the sum up to n=2 (1/2 + 2/3 ≈ 0.5 + 0.666... ≈ 1.166...).

Similarly, for N=5:

Sum = 2 - 1/F_6 - 1/F_7 = 2 - 1/8 - 1/13 ≈ 2 - 0.125 - 0.076923 ≈ 2 - 0.201923 ≈ 1.79807... which should match the sum up to n=5. Let's compute the original sum up to n=5:

1/2 + 2/3 + 3/10 + 5/24 + 8/65 ≈ 0.5 + 0.6667 + 0.3 + 0.2083 + 0.1231 ≈ 0.5 + 0.6667 = 1.1667; 1.1667 + 0.3 = 1.4667; 1.4667 + 0.2083 ≈ 1.675; 1.675 + 0.1231 ≈ 1.7981. Yes, that's approximately 1.7981, which matches 2 - 1/8 - 1/13 ≈ 1.79807...

Therefore, the formula seems correct. Thus, as N approaches infinity, F_{N+1} and F_{N+2} approach infinity, so 1/F_{N+1} and 1/F_{N+2} approach zero, hence the sum converges to 2. Therefore, the value of the infinite series is 2.
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
Wait, but let me check if this aligns with the initial terms.

<prompt_6.1> Confirm the result by checking if the derived formula aligns with the initial terms and compute the limit as \(N\) approaches infinity. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
Wait, when N approaches infinity, F_{N+1} and F_{N+2} become very large, so their reciprocals go to zero, so the sum approaches 2 - 0 - 0 = 2. Therefore, the sum of the series is 2.

But let me just verify once again with another approach to ensure there's no mistake. Let's consider the general term:

\(\frac{F_{n+1}}{F_n F_{n+2}} = \frac{1}{F_n} - \frac{1}{F_{n+2}}\)

Therefore, summing from n=1 to ∞:

Sum = Σ_{n=1}^∞ (1/F_n - 1/F_{n+2}) = (Σ_{n=1}^∞ 1/F_n) - (Σ_{n=1}^∞ 1/F_{n+2})

But the second sum is just Σ_{k=3}^∞ 1/F_k, where k = n + 2. Therefore, the total sum is:

(Σ_{n=1}^∞ 1/F_n) - (Σ_{k=3}^∞ 1/F_k) = 1/F_1 + 1/F_2 = 1 + 1 = 2

Because the sums from n=3 to infinity cancel out. That's a straightforward way to see it. Therefore, the total sum is indeed 2.
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
**Final Answer**
\boxed{2}
</Plan_7>

<Execution_7>
No prompt to execute.
</Execution_7>

<Final_answer>
Therefore, the sum of the series is \(\boxed{2}\).
</Final_answer>
</think>