# Alternating series and reciprocal Fibonacci constant

We know that reciprocal Fibonacci constant $$\sum_{n=1}^{\infty} \frac{1}{F_n} = \frac{1}{1} + \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{8} + \frac{1}{13} + \frac{1}{21} + \cdots \approx 3.3598856662 \dots .$$

Evaluate: $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{F_n}$$

I got to say after spending some time with this problem: that's a truly remarkable series, I would love if other people could look into this problem. This is not a full answer, this was just my way to approach the question and I couldn't find some "nice and short" closed-form of the series. Let's begin!

Notice how the series can be split up into an odd and an even part:

$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{F_n} = \sum_{n=0}^{\infty} \frac{1}{F_{2n+1}}-\sum_{n=1}^{\infty}\frac{1}{F_{2n}}$$

Let's first evaluate the series $$\sum_{n=0}^{\infty} \frac{1}{F_{2n}}$$ because I feel like that's easier a little bit easier.

Evaluating $$\sum_{n=0}^{\infty} \frac{1}{F_{2n}}$$

This is quite a tricky process, but let's try to work it through. Let $$\phi$$ denote the golden ratio.

\begin{align} \sum_{n=1}^{\infty} \frac{1}{F_{2n}} &= \sqrt{5}\sum_{n=1}^\infty \frac{\phi^{2n}}{\phi^{4n}-1} \\\\ & = \sqrt{5}\sum_{n=1}^{\infty}\frac{1}{\phi^{2n}-1}-\frac{1}{\phi^{4n}-1} \\\\ & = \sqrt{5}\left(\sum_{n=1}^{\infty}\frac{1}{\phi^{2n}-1}-\sum_{n=1}^{\infty}\frac{1}{\phi^{4n}-1}\right)\end{align}

The sums take form of what's known as a Lambert series. A Lambert series is defined as:

$$F(x)=\sum_{n=1}^\infty a_n\frac{x^n}{1-x^n}$$

Now let the case when $$a_n = 1$$ be denoted by $$L(x)$$:

$$L(x) = \sum_{n=1}^\infty \frac{x^n}{1-x^n} = \sum_{n=1}^\infty \frac{1}{x^{-n}-1}$$

And so in the case above we have:

$$\sum_{n=1}^{\infty} \frac{1}{F_{2n}} = \sqrt{5}\left(L(\phi^{-2})-L(\phi^{-4})\right)$$

Now, it is known (and I don't know how; I can't prove it...) that:

$$L(x) = \frac{\psi_x(1)+\ln(1-x)}{\ln(x)}$$

Here $$\psi_x(1)$$ denotes the q-Polygamma function. Using the identity above we get:

$$\sqrt{5}\left(L(\phi^{-2})-L(\phi^{-4})\right) = \frac{\sqrt{5}}{8\ln(\phi)}\left(\ln(5)+2\psi_{\phi^{-4}}(1)-4\psi_{\phi^{-2}}(1)\right)$$

This as far as I can go. Thus we have arrived at the result that:

$$\sum_{n=1}^{\infty} \frac{1}{F_{2n}} = \frac{\sqrt{5}}{8\ln(\phi)}\left(\ln(5)+2\psi_{\phi^{-4}}(1)-4\psi_{\phi^{-2}}(1)\right)$$

Now according to WolframMathWorld:

$$\frac{\sqrt{5}}{8\ln(\phi)}\left(\ln(5)+2\psi_{\phi^{-4}}(1)-4\psi_{\phi^{-2}}(1)\right) = \frac{\sqrt{5}}{4\ln(\phi)}\left(\psi_{\phi^2}\left(1-\frac{i\pi}{2\ln(\phi)}\right)-\psi_{\phi^2}(1)+i\pi\right)$$

But I have know idea, how they got there. Anyways, the value equals approximately $$1.5353705...$$

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

This derivation is pretty much the same as the previous one; you express the Fibonacci numbers in terms of the golden ratio and the simplify it into a Lambert series and then express it in terms of the q-Polygamma function. I did part of this derivation independently, but found that all of these are found at WolframMathWorld. You'll end up with a similar expression to the sum of even Fibonacci numbers:

$$\sum_{n=0}^{\infty} \frac{1}{F_{2n+1}} = -\frac{\sqrt{5}}{4\ln(\phi)}\left(\pi-i\left(\psi_{\phi^2}\left(\frac{1}{2}-\frac{i\pi}{4\ln(\phi)}\right)-\psi_{\phi^2}\left(\frac{1}{2}+\frac{i\pi}{4\ln(\phi)}\right)\right)\right) \tag{1}$$

According to WolframMathWorld an identity (I have no idea how to prove this) is that:

$$\left(\pi-i\left(\psi_{\phi^2}\left(\frac{1}{2}-\frac{i\pi}{4\ln(\phi)}\right)-\psi_{\phi^2}\left(\frac{1}{2}+\frac{i\pi}{4\ln(\phi)}\right)\right)\right) = -(\ln\phi)\vartheta_2^2(\phi^{-2})$$

where $$\vartheta_2(q)$$ is an Jacobi theta function (read more here). So using this identity on equation $$(1)$$ we get:

$$-\frac{\sqrt{5}}{4\ln(\phi)}\left(\pi-i\left(\psi_{\phi^2}\left(\frac{1}{2}-\frac{i\pi}{4\ln(\phi)}-\psi_{\phi^2}(\frac{1}{2}+\frac{i\pi}{4\ln(\phi)}\right)\right)\right) = \frac{\sqrt{5}}{4}\vartheta_2^2(\phi^{-2})$$

Which is quite a big simplification if you ask me.

Puting everything together

Recall that: $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{F_n} = \sum_{n=0}^{\infty} \frac{1}{F_{2n+1}}-\sum_{n=1}^{\infty}\frac{1}{F_{2n}}$$

Using the results we have established we get:

$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{F_n} = -\frac{\sqrt{5}}{4\ln(\phi)}\left(\pi-i\left(\psi_{\phi^2}(\frac{1}{2}-\frac{i\pi}{4\ln(\phi)}-\psi_{\phi^2}(\frac{1}{2}+\frac{i\pi}{4\ln(\phi)}\right)\right) - \frac{\sqrt{5}}{4\ln(\phi)}\left(\psi_{\phi^2}\left(1-\frac{i\pi}{2\ln(\phi)}\right)-\psi_{\phi^2}(1)+i\pi\right)$$

\begin{align} &= -\frac{\sqrt{5}}{4\ln(\phi)} \left(\pi-i\left(\psi_{\phi^2}\left(\frac{1}{2}-\frac{i\pi}{4\ln(\phi)}\right)-\psi_{\phi^2}\left(\frac{1}{2}+\frac{i\pi}{4\ln(\phi)}\right)\right) + \psi_{\phi^2}\left(1-\frac{i\pi}{2\ln(\phi)}\right)-\psi_{\phi^2}(1)+i\pi\right)\end{align}

Now let:

$$\Omega = \frac{i\pi}{4\ln(\phi)}$$

Therefore we get:

\begin{align} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{F_n} &= -\frac{\sqrt{5}}{4\ln(\phi)} \left(\pi-i\left(\psi_{\phi^2}\left(\frac{1}{2}-\Omega\right)-\psi_{\phi^2}\left(\frac{1}{2}+\Omega\right)\right) + \psi_{\phi^2}\left(1-2\Omega\right)-\psi_{\phi^2}(1)+i\pi\right)\end{align}

That's pretty horrible looking, but I don't know how to simplify it :(

Also note that if you want the series in terms of the Reciprocal Fibonacci Constant $$\psi$$,

$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{F_n} = \sqrt{\psi^2-4F}$$

where $$F$$ is:

$$F = \left(\sum_{n=0}^{\infty} \frac{1}{F_{2n+1}}\right)\left(\sum_{n=1}^{\infty}\frac{1}{F_{2n}}\right)$$