A closed form for the sum $\frac{a}{b}+\frac{a\cdot(a+1)}{b\cdot(b+1)}+\frac{a\cdot(a+1)\cdot(a+2)}{b\cdot(b+1)\cdot(b+2)}+\cdots$

This identity is easy to deduce once you notice that

$$\frac1{\binom nk}-\frac1{\binom{n+1}k}=\frac k{k+1}\frac1{\binom{n+1}{k+1}}$$

It thus follows that

$$\sum_{n=k}^\infty\frac1{\binom nk}=\frac k{k-1}\sum_{n=k}^\infty\left(\frac1{\binom{n-1}{k-1}}-\frac1{\binom n{k-1}}\right)=\frac k{k-1}\frac1{\binom{k-1}{k-1}}=\frac k{k-1}$$

and better yet,

$$\sum_{n=0}^\infty\frac1{\binom{b+n}{b-a}}=\frac{b-a}{b-a+1}\sum_{n=0}^\infty\left(\frac1{\binom{b+n-1}{b-a-1}}-\frac1{\binom{b+n}{b-a-1}}\right)=\frac{b-a}{b-a+1}\frac1{\binom{b-1}{b-a-1}}$$

where the binomial expectedly cancels near the beginning of your calculations.

Euler is your friend. There is Gauss' Hypergeometric function (defined by Euler, that guy Euler was robbed, there isn't enough named after him):

$${}_2 F_{1}(a,b;c;z) = 1 + \frac{a b z}{c} + \frac{a(a+1) b(b+1) z^2}{c(c+1) 2!} + \frac{a(a+1)(a+2) b(b+1)(b+2) z^3}{c(c+1)(c+2) 3!} + \ldots $$

and you are asking about the value of

$${}_2 F_{1}(a,1;c;1) - 1.$$

But there is the simple formula (due to Euler)

$${}_2 F_{1}(a,b;c;1) = \frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(c-a) \Gamma(c - b)}$$

You can prove this from the more general integral representation $${}_2 F_{1}(a,b;c;z) = \frac{\Gamma(c) \Gamma(b)}{\Gamma(c-b) } \int^{1}_{0} t^{b-1} (1-t)^{c-b-1} (1 - t z)^{-a} dz$$

which follows by expanding out the last term and applying Euler's beta integral. In particular, using basic properties of the Gamma function you find that

$${}_2 F_{1}(a,1;c;1) - 1 = \frac{a}{c-a-1}$$

For example, with $a = 17$, and $c = 76$, and then dividing the answer by $75$, you get

$$\frac{17}{75 \cdot 76} + \frac{17 \cdot 18}{75 \cdot 76 \cdot 77} + \ldots = \frac{1}{75} \cdot \frac{17}{76 - 17 - 1} = \frac{17}{4350}.$$

The sum in question can actually be evaluated in quite an elementary way as follows $$\begin{align} \frac{a}{b}+\frac{a\cdot(a+1)}{b\cdot(b+1)}+\frac{a\cdot(a+1)\cdot(a+2)}{b\cdot(b+1)\cdot(b+2)}+\cdots &=\frac{(b-1)!}{(a-1)!}\sum_{n=0}^\infty\frac{(a+n)!}{(b+n)!}\\ &=\frac{(b-1)!}{(a-1)!}\sum_{n=0}^\infty\frac1{(n+a+1)\cdots(n+b)}\\ &=\frac{(b-1)!}{(a-1)!}\sum_{n=0}^\infty\frac{\frac1{(n+a+1)(n+b)}}{(n+a+2)\cdots(n+b-1)}\\ &=\frac{(b-1)!}{(a-1)!}\sum_{n=0}^\infty\frac{\frac1{b-a-1}\left(\frac1{n+a+1}-\frac1{n+b}\right)}{(n+a+2)\cdots(n+b-1)}\\ &=\frac{(b-1)!}{(a-1)!\cdot(b-a-1)}\sum_{n=0}^\infty\left(\frac1{(n+a+1)\cdots(n+b-1)}-\frac1{(n+a+2)\cdots(n+b)}\right)\\ &=\frac{(b-1)!}{(a-1)!\cdot(b-a-1)}\left(\frac1{(a+1)\cdots(b-1)}\right)\\ &=\frac{(b-1)!}{(a-1)!\cdot(b-a-1)}\left(\frac{a!}{(b-1)!}\right)\\ &=\boxed{\frac{a}{b-a-1}}\\ \end{align}$$

Also, using the methods found in this paper, we can prove the following additional result $$\begin{align} \sum_{n=k}^\infty\frac1{\binom{n}{k}} &=\sum_{n=0}^\infty\frac1{\binom{n+k}{k}}\\ &=\sum_{n=0}^\infty\frac{n!\cdot k!}{(n+k)!}\\ &=k\sum_{n=0}^\infty\frac{n!\cdot (k-1)!}{(n+k)!}\\ &=k\sum_{n=0}^\infty B(n+1,k)\\ &=k\sum_{n=0}^\infty \int_0^1 t^n (1-t)^{k-1}\mathrm{d}t\\ &=k\int_0^1(1-t)^{k-1}\left(\sum_{n=0}^\infty t^n\right)\mathrm{d}t\\ &=k\int_0^1(1-t)^{k-2}\mathrm{d}t\\ &=\boxed{\frac{k}{k-1}}\\ \end{align}$$