Let

Sk=∞∑n=111+n21+1⋱1+11+n2,k rows in the continued fraction

So for example, the terms of the sum S6 are

11+n21+11+11+11+11+n2

Using a symbolic computation software (Mathematica), I got the following interesting results:

S4=π4(coth(π)+√3coth(√3π))−12S6=π4(√2coth(√2π)+√43coth(√43π))−12S8=π4(√32coth(√32π)+√74coth(√74π))−12S10=π4(√53coth(√53π)+√117coth(√117π))−12S12=π4(√85coth(√85π)+√1811coth(√117π))−12S14=π4(√138coth(√138π)+√2918coth(√2918π))−12.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.

Hence:

Conjecture:S2k=π4(√FkFk−1coth(√FkFk−1π)+√LkLk−1coth(√LkLk−1π))−12I 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

limk→∞S2k=√φπcoth(√φπ)−12

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

**Answer**

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

11+n21+11+⋱1+11+11+n2=Fk−1n2+FkFk−2n4+2Fk−1n2+Fk

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.

1Fk−2n4+2Fk−1n2+Fk=1(F(k−2)/2n2+Fk/2)(L(k−2)/2n2+Lk/2)

This follows from the fact that

Fn+Fn+2=Ln+1

Now we need the complex roots of each term. On the left hand side, these are

±i√Fk/2F(k−2)/2

And on the right side these are just

±i√Lk/2L(k−2)/2

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.

a1=√Fk/2F(k−2)/2

a2=√Lk/2L(k−2)/2

If we plug these roots back into the equation from the first step, we get an equation that looks like this

f(n)=Fk−1n2+Fk(n+ia1)(n−ia1)(n+ia2)(n−ia2)

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).

ba1=π(Fk−1a21−Fk)coth(πa1)2a1(ia1+ia2)(ia1−ia2)=−π(Fk−1a21−Fk)coth(πa1)2a1(a21−a22)

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.

\sum_{n=-\infty}^{\infty}\frac{F_{k-1}\:n^2+F_{k}}{(n^2+a_1^2)(n^2+a_2^2)}=-(2b_{a1}+2b_{a2})=\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

-1-\frac{F_{k-1}\:\frac{F_{k/2}}{F_{k/2-1}}-F_{k}}{\frac{L_{k/2}}{L_{k/2-1}}}=-\frac{2}{F_{k-1}+1}

and similarly

1-\frac{F_{k-1}\:\frac{L_{k/2}}{L_{k/2-1}}-F_{k}}{\frac{F_{k/2}}{F_{k/2-1}}}=-\frac{2F_{k-1}}{F_{k-1}-1}

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 🙂

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