Closed form solution for ∑∞n=111+n21+1⋱1+11+n2\sum_{n=1}^\infty\frac{1}{1+\frac{n^2}{1+\frac{1}{\stackrel{\ddots}{1+\frac{1}{1+n^2}}}}}.

Sk=n=111+n21+11+11+n2,k rows in the continued fraction
So for example, the terms of the sum S6 are
Using a symbolic computation software (Mathematica), I got the following interesting results:

The numbers appearing at the first coth term are easy to guess: they are the famous Fibonacci numbers.

The numbers at the second coth term can also be guessed: they appear to be the Lucas numbers. Those are constructed like the Fibonacci numbers but starting with 2,1 instead of 0,1.


Conjecture: S2k=π4(FkFk1coth(FkFk1π)+LkLk1coth(LkLk1π))12

I have verified this conjecture for many k‘s and it always work out perfectly. To me this is quite amazing, but I am not able to verify the conjecture. Can anyone prove it?

Moreover, if true the conjecture implies that


which is also very nice (π and φ don’t meet very often).


This is again only a part of a proof, because I believe I made a mistake somewhere along the line. Anyway, we’ll start from this equation
where k is equal to the number of horizontal lines on the left hand side. user109899 went off of the same premise, but we’ll change it up a little. Let’s start by partially factoring the denominator on the right hand side.
This follows from the fact that
Now we need the complex roots of each term. On the left hand side, these are
And on the right side these are just
At this point I’m also going to define a couple of variables to represent these roots in a way that we can more easily work with.
If we plug these roots back into the equation from the first step, we get an equation that looks like this
For a more detailed description of this next step, read [this][1] thread. I won’t explain this much since it’s already shown there, but we need to find the residues of an equation g(z):=πcot(πz)f(n) with poles at ±ia1 and ±ia2 (Note that the residue for each pole is identical to its opposite pole because coth is an odd function).

b_{a2}=\frac{\pi(F_{k-1}\:a_2^2-F_{k})\:\coth(\pi a_2)}{2a_2(ia_2+ia_1)(ia_2-ia_1)}=-\frac{\pi(F_{k-1}\:a_2^2-F_{k})\:\coth(\pi a_2)}{2a_2(a_2^2-a_1^2)}

We can then plug f(x) into an infinite sum to evaluate it based upon these residues.

\frac{\pi(F_{k-1}\:a_1^2-F_{k})\:\coth(\pi a_1)}{a_1(a_1^2-a_2^2)}+\frac{\pi(F_{k-1}\:a_2^2-F_{k})\:\coth(\pi a_2)}{a_2(a_2^2-a_1^2)}

Using the identity F_k-F_{k+1}\frac{F_{k/2}}{F_{(k+1)/2}}=-1 and F_k-F_{k+1}\frac{L_{k/2}}{L_{(k+1)/2}}=1 for odd k, we can simplify this a bit further.

\small{\pi\left (F_{k-1}-\frac{F_{k}}{a_1^2}-\frac{F_{k-1}\:a_1^2-F_{k}}{a_2^2}\right )\:a_1\coth(\pi a_1)+\pi\left (F_{k-1}-\frac{F_{k}}{a_2^2}-\frac{F_{k-1}\:a_2^2-F_{k}}{a_1^2}\right )\:a_2\coth(\pi a_2)=}

\pi\left (\left (-1-\frac{F_{k-1}\:a_1^2-F_{k}}{a_2^2}\right )a_1\coth(\pi a_1)+\left (1-\frac{F_{k-1}\:a_2^2-F_{k}}{a_1^2}\right )a_2\coth(\pi a_2)\right )

Because the original function is symmetric for negative n, we can subtract the zero term and divide the result by two in order to get the sum from 1 to infinity.

\frac{F_{k-1}\:0^2+F_{k}}{F_{k-2}\:0^4+2F_{k-1}\:0^2+F_{k}} = 1

\small{\sum_{n=1}^{\infty}\frac{F_{k-1}\:n^2+F_{k}}{(n^2+a_1^2)(n^2+a_2^2)}=\frac{\pi\left (\left (-1-\frac{F_{k-1}\:a_1^2-F_{k}}{a_2^2}\right )a_1\coth(\pi a_1)+\left (1-\frac{F_{k-1}\:a_2^2-F_{k}}{a_1^2}\right )a_2\coth(\pi a_2)\right ) – 1}{2}}=

\small{\frac{\pi}{2}\left (\left (-1-\frac{F_{k-1}\:a_1^2-F_{k}}{a_2^2}\right )a_1\coth(\pi a_1)+\left (1-\frac{F_{k-1}\:a_2^2-F_{k}}{a_1^2}\right )a_2\coth(\pi a_2)\right ) – \frac{1}{2}}

Not-so-simple algebra and applying certain properties of fibonacci numbers shows that

and similarly

Plugging these values into the equation now gives

\frac{\pi}{2}\left (\left (-\frac{2}{F_{k-1}+1}\right )a_1\coth(\pi a_1)+\left (-\frac{2F_{k-1}}{F_{k-1}-1}\right )a_2\coth(\pi a_2)\right ) – \frac{1}{2}

At this point, if we expand the variables, we get

\small{\frac{\pi}{2}\left (\left (-\frac{2}{F_{k-1}+1}\right )\sqrt{\frac{F_{k/2}}{F_{k/2-1}}}\coth\left (\sqrt{\frac{F_{k/2}}{F_{k/2-1}}}\pi \right )+\left (-\frac{2F_{k-1}}{F_{k-1}-1}\right )\sqrt{\frac{L_{k/2}}{L_{k/2-1}}}\coth\left (\sqrt{\frac{L_{k/2}}{L_{k/2-1}}}\pi \right )\right ) – \frac{1}{2}}

Which is very similar to your original equation, but is not the same. I think I must have made a mistake in my math somewhere, but hopefully this helps in finding a correct proof. I’ve spent over half a day on this equation now, though, so I’ll leave that to someone else 🙂

Source : Link , Question Author : Simon Parker , Answer Author : Ethan MacBrough

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