The Fibonacci Array

Denoting by $F_n$ , the $n$ th Fibonacci number , note that we can write the sum as $S=\sum_{k\ge 2}\sum_{n\ge 2}\frac{F_n}{k^n}\\=\sum_{k\ge 2}\sum_{n\ge 2}\frac{\phi^n-(1-\phi)^n}{2\phi-1}$ where $\phi=\frac{1+\sqrt{5}}{2}$ , is the golden ratio. Then, noting that $1<\phi<2$ , we can simplify the sum as $(2\phi-1)S=\sum_{n\ge 2}\sum_{k\ge 2}\left(\frac{\phi^n}{k^n}-\frac{(1-\phi)^n}{k^n}\right)\\=\sum_{k\ge 2}\left(\frac{(\phi/k)^2}{1-\phi/k}-\frac{((1-\phi)/k)^2}{1-(1-\phi)/k}\right)\\=\phi\sum_{k\ge 2}\left(\frac{1}{k-\phi}-\frac{1}{k}\right)-(1-\phi)\sum_{k\ge 2}\left(\frac{1}{k-1+\phi}-\frac{1}{k}\right)$ Now, the digamma function satisfies the following property: $\psi(x+1)=-\gamma+\sum_{k\ge 1}\left(\frac{1}{k}-\frac{1}{x+k}\right)$ where $\gamma$ is the Euler-Mascheroni constant. Then, it follows after a little bit of mainipulation, $S=\frac{(1-\phi)\psi(\phi)-\phi\psi(1-\phi)}{2\phi-1}-\gamma+2$ We can further simplify the expression using the reflection formula satisfied by $\psi(\cdot)$ $\psi(1-x)-\psi(x)=\pi\cot(\pi x)$ to obtain $S=2-\gamma+\frac{\pi(\phi-1)\cot (\pi \phi)}{2\phi-1}-\psi(1-\phi)$ which evaluates to $\approx 2.16522$ . Thus the answer is $\boxed{216522}$ .

Define $S$ such that

$S = \frac{1}{x} + \frac{1}{x^2} + \frac {2}{x^3} + \frac{3}{x^4} + \frac {5}{x^5} + ...$

so

$\frac{S}{x} = \frac{1}{x^2} + \frac{1}{x^3} + \frac {2}{x^4} + \frac{3}{x^5} + \frac {5}{x^6} + ...$

Subtract the two and we get

$S - \frac {S}{x} = \frac{1}{x} + \frac{1}{x^2} ( \frac{1}{x} + \frac{1}{x^2} + \frac {2}{x^3} + \frac{3}{x^4} + \frac {5}{x^5} + ... )$

$S - \frac {S}{x} = \frac{1}{x} + \frac{S}{x^2}$

Multiplying $x^2$ on both sides,

$x^2S - xS = x + S$

so

$S = \frac{x}{x^2 - x - 1}$

The array above can now be condensed to

$\large \sum_{x=1}^{\infty} \frac{x}{x^2 - x - 1}$

which, by the way, diverges. Note: I'm sorry I cannot give an in depth explanation to this as I am not very well versed with the different series convergence tests available. However, on an intuitive notion, we can see that the first row (the one in red) consists of plainly infinite 1s (which means this will blow up to infinity), while the first column is the famous divergent series $\sum_{n=1}^{\infty} \frac{1}{n}$ . Since these two take part in the array in question, then it can be said that this array series indeed diverges.

So, to make for adjustments by removing the divergent parts, we get

$\large \sum_{x=2}^{\infty} \frac{x}{x^2 - x - 1} - \frac{1}{x}$

We now start at $x =2$ because the first row (where $x=1$ ) is removed, and the subtracted expression takes into account the removal of the first column.

I'll stop here for now.