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.
Best Answer
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.