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

I watched this YouTube video that calculates the sum $$\frac{1}{3\cdot4}+\frac{1\cdot2}{3\cdot4\cdot5}+\frac{1\cdot2\cdot3}{3\cdot4\cdot5\cdot6}+\cdots=\frac16$$ then they ask, as a challenge to the viewer, what is the value of the sum $$\frac{17}{75\cdot76}+\frac{17\cdot18}{75\cdot76\cdot77}+\frac{17\cdot18\cdot19}{75\cdot76\cdot77\cdot78}+\cdots$$ This got me thinking about a way to generalise this type of sum, i.e. how can one calculate the value of the sum $$\frac{a}{b}+\frac{a\cdot(a+1)}{b\cdot(b+1)}+\frac{a\cdot(a+1)\cdot(a+2)}{b\cdot(b+1)\cdot(b+2)}+\cdots$$ where $a,b\in\mathbb{N}$ and $a\lt b$ . We can rewrite this sum as $$\begin{align} \frac{(b-1)!}{(a-1)!}\sum_{n=0}^\infty\frac{(a+n)!}{(b+n)!} &=\frac{(b-1)!}{(a-1)!\cdot(b-a)!}\sum_{n=0}^\infty\frac{(a+n)!\cdot(b-a)!}{(b+n)!}\\ &=\frac{(b-1)!}{(a-1)!\cdot(b-a)!}\sum_{n=0}^\infty\frac1{\binom{b+n}{b-a}}\\ &=\frac{(b-1)!}{(a-1)!\cdot(b-a)!}\left(\sum_{n=b-a}^\infty\frac1{\binom{n}{b-a}}-\sum_{n=b-a}^{b-1}\frac1{\binom{n}{b-a}}\right)\\ \end{align}$$ So this effectively simplifies down to the following problem:

How can we evaluate the sum $$\sum_{n=k}^\infty \frac1{\binom{n}{k}}$$ for $k\in\mathbb{N}\setminus\{1\}$ in a closed form?

Numerically it appears that the solution is $$\boxed{\sum_{n=k}^\infty \frac1{\binom{n}{k}}=\frac{k}{k-1}}$$ which would mean that a closed form for our sum is $$\boxed{\frac{a}{b}+\frac{a\cdot(a+1)}{b\cdot(b+1)}+\frac{a\cdot(a+1)\cdot(a+2)}{b\cdot(b+1)\cdot(b+2)}+\cdots=\frac{(b-1)!}{(a-1)!\cdot(b-a)!}\left(\frac{b-a}{b-a-1}-\sum_{n=b-a}^{b-1}\frac1{\binom{n}{b-a}}\right)}$$ testing this solution for our example gives $$\begin{align} \frac{17}{75\cdot76}+\frac{17\cdot18}{75\cdot76\cdot77}+\frac{17\cdot18\cdot19}{75\cdot76\cdot77\cdot78}+\cdots &=\frac1{75}\left(\frac{17}{76}+\frac{17\cdot18}{76\cdot77}+\frac{17\cdot18\cdot19}{76\cdot77\cdot78}+\cdots\right)\\ &=\frac1{75}\left(\frac{(76-1)!}{(17-1)!\cdot(76-17)!}\left(\frac{76-17}{76-17-1}-\sum_{n=76-17}^{76-1}\frac1{\binom{n}{76-17}}\right)\right)\\ &=114000634335804\left(\frac{59}{58}-\sum_{n=59}^{75}\frac1{\binom{n}{59}}\right)\\ &=114000634335804\left(\frac{59}{58}-\frac{1023230845711831}{1005887950021800}\right)\\ &=114000634335804\left(\frac1{29170750550632200}\right)\\ &=\frac{17}{4350}\\ \end{align}$$ which seems to agree with numerical evaluation, but how do I prove this result?

Edit: There is actually a much better closed form for this result as follows $$\boxed{\frac{a}{b}+\frac{a\cdot(a+1)}{b\cdot(b+1)}+\frac{a\cdot(a+1)\cdot(a+2)}{b\cdot(b+1)\cdot(b+2)}+\cdots=\frac{a}{b-a-1}}$$ which is found in the supplied answers.

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

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