Impossible sum.

I'll assume $0\not\in\mathbb N$ (there are a lot of different definitions, hence I've pointed it out).

Fact 1:

$\displaystyle f(k,1)=k,\:\forall k\in\mathbb N$

Proof by induction:

Notice that $f(1,1)=1$ .

Assume our statement is true when $f(k,1)=k$ , where $k\in\mathbb N$ .

We'll prove that $f(k+1,1)=k+1$ .

Whenever $m=1, n=k$ , we have $\displaystyle\frac{f(k+1,1)}{f(k,1)}=\frac{k+1}{k}\iff f(k+1,1)=k+1\square$

Fact 2:

$\displaystyle f(m,m+1)=m(m+1),\:\forall m\in\mathbb N$

Proof:

Whenever $n=1$ , we have \displaystyle \frac{f(m,m+1)}{f(m,1)}=m+1\stackrel{\text{Fact 1}}\iff f(m,m+1)=m(m+1), \:\forall m\in\mathbb N\square

Fact 2 lets us conclude that

$\displaystyle S=\frac{1}{f(1,2)f(3,4)}+\frac{1}{f(2,3)f(4,5)}+\cdots$ $\displaystyle =\frac{1}{1\cdot 2\cdot 3\cdot 4}+\frac{1}{2\cdot 3\cdot 4\cdot 5}+\cdots$ $\displaystyle =\frac{0!}{4!}+\frac{1!}{5!}+\frac{2!}{6!}+\cdots$ $\displaystyle =\frac{1}{18}=0.05555...\approx \boxed{0.056}$

The latter summation was calculated with some help from Wolfram Alpha .

Note that $f(m,n)=\frac{m}{m-n}f(n,m-n)$

Let $g(m,n)=\frac{mn}{f(m,n)}$

Hence, 1. $g(n,n)=n$ , $g(m,n)$ $=$ $g(n,m)$ , $g(m,n)=\frac{mn}{f(m,n)}=\frac{mn}{\frac{m}{m-n}f(n,m-n)}=g(n,m-n)$

Now use Euclidean algorithm to show $g(m,n)=gcd(m,n)$

So, $f(m,n)=lcm(m,n)$

$S=\sum_{r=1}^{∞}\frac{1}{f(r,r+1)f(r+2,r+3)}$

Hence, $S=\sum_{r=1}^{∞}\frac{1}{r(r+1)(r+2)(r+3)}$

For $n$ terms, $S_{n}=C-\frac{1}{3(n+1)(n+2)(n+3)}$ where $C$ is a constant.

Letting $n=1$ , $C=\frac{1}{18}$ .

So, $S=\frac{1}{18}$