Have you grabbed all the properties of Binomial Coefficients?

$\displaystyle \sum_{r=0}^{n}{{(-1)}^{r+1} \frac{\binom{n}{r}}{n+r}}$

So, this is our sum. Consider this first:

$\displaystyle \sum_{r=0}^{n}{\binom{n}{r} {x}^{r}} = {(1+x)}^{n}$

$\displaystyle \sum_{r=0}^{n}{\binom{n}{r} {x}^{r+n-1}} = {(1+x)}^{n}{x}^{n-1}$

Now, integrate both sides w.r.t. x from 0 to -1,

$\displaystyle \sum_{r=0}^{n}{\frac{\binom{n}{r}}{n+r} {(-1)}^{r+n}} = \int_{0}^{-1}{{(1+x)}^{n}{x}^{n-1}}$

In the integral in RHS, substitute $x=-u$

$\int_{0}^{1}{{(1-u)}^{n}{(-1)}^{n}{u}^{n-1}}$

$\displaystyle = {(-1)}^{n}\int_{0}^{1}{{(1-u)}^{n}{u}^{n-1}}$

It looks familiar, right? Beta function!

$\displaystyle = {(-1)}^{n}\frac{n!(n-1)!}{(2n)!}$

Now, we need to find n.

$\displaystyle \sum_{a=1}^{\infty}{\frac{{20}^{a}}{({5}^{a}-{4}^{a})({5}^{a+1}-{4}^{a+1})}}$

Telescoping must be the first guess, and that's right!

$\displaystyle \sum_{a=1}^{\infty}{\frac{{4}^{a}}{{5}^{a}-{4}^{a}} -\frac{{4}^{a+1}}{{5}^{a+1}-{4}^{a+1}}}$

which telescopes to $\displaystyle 4$ , $\displaystyle n = 2 \times 4 = 8$

Substitute this in our sum formula,

$\displaystyle {(-1)}^{8}\frac{8!(8-1)!}{(2*8)!} = \boxed{\frac{1}{102960}}$

It's really a good question. I like it. I think it should be level 5.