Showing that $\sum_{k=0}^{n}{m+k\choose m}^{-1}=\frac{m}{m-1}\left[1-{m+n\choose m-1}^{-1}\right],m>1.$ Let  $$S_{m,n}=\sum_{k=0}^{n}{m+k\choose m}^{-1}$$
Then $S_{0,n}=n+1, S_{1,n}=H_{n+1}.$
For $m>1$, let us consider the integral $$I_{m,k}=\int_{0}^{1} (1-x^{1/m})^k dx$$
which by using $x=\sin^{2m} t$ and Beta function, can be expressed as
$$I_{m,k}=2m\int_{0}^{\frac{\pi}{2}}\sin^{2m-1} t ~ \cos^{2k+1} t~ dt= \frac{\Gamma(m+1) \Gamma(k+1)}{\Gamma(m+k+1)}={m+k\choose m}^{-1}.$$
Next, we can write $$S_{m,n}=\sum_{k=0}^{n} I_{m,k}=\int_{0}^{1}dx \sum_{k=0}^n (1-x^{1/m})^k=\int_{0}^{1} x^{-1/m}[1-(1-x^{1/m})^{n+1}]~ dx.$$
$$\implies S_{m,n}=\frac{m}{m-1}-\int_0^1 x^{-1/m}(1-x^{1/m})^{n+1} dx$$
Again by using $x=\sin^{2m} t$ and Beta-integral, we get
$$S_{m.n}=\frac{m}{m-1}-2m\int_{0}^{\frac{\pi}{2}} \sin^{2m-3}~\cos^{2n+3} ~dt=\frac{m}{m-1}-\frac{\Gamma(m)\Gamma(n+2)}{\Gamma(m+m+1)}. $$
Upon simplification we have $$S_{m,n}=\frac{m}{m-1}\left[1-{m+n\choose m-1}^{-1}\right],m>1.$$
The question us how else this result can be obtained?
 A: We seek to show that for $m\gt 1$
$$S_{m,n} = \sum_{k=0}^n {m+k\choose m}^{-1}
= \frac{m}{m-1}
\left[1-{m+n\choose m-1}^{-1}\right].$$
We have for the LHS using an Iverson bracket:
$$[w^n] \frac{1}{1-w} 
\sum_{k\ge 0} {m+k\choose m}^{-1} w^k.$$
Recall from MSE
4316307
that with $1\le k\le n$
$${n\choose k}^{-1}
= k [z^n] \log\frac{1}{1-z}
(z-1)^{n-k}.$$
We get with $m\ge 1$ as per requirement on $k$
$$m \; \underset{z}{\mathrm{res}} \; 
\frac{1}{z^{m+1}} \log\frac{1}{1-z} 
[w^n] \frac{1}{1-w} 
\sum_{k\ge 0} w^k z^{-k} (z-1)^k
\\ = m \; \underset{z}{\mathrm{res}} \; 
\frac{1}{z^{m+1}} \log\frac{1}{1-z} 
[w^n] \frac{1}{1-w} \frac{1}{1-w(z-1)/z}
\\ = m  \; \underset{z}{\mathrm{res}} \; 
\frac{1}{z^{m}} \log\frac{1}{1-z}
\; \underset{w}{\mathrm{res}} \;  
\frac{1}{w^{n+1}} \frac{1}{1-w} 
\frac{1}{z-w(z-1)}.$$
Now residues sum to zero and the residue at infinity in $w$ is zero by
inspection, so we may evaluate by taking minus the residue at $w=1$ and
minus the residue at $w=z/(z-1).$ For $w=1$ start by writing
$$- m  \; \underset{z}{\mathrm{res}} \; 
\frac{1}{z^{m}} \log\frac{1}{1-z}
\; \underset{w}{\mathrm{res}} \;  
\frac{1}{w^{n+1}} \frac{1}{w-1} 
\frac{1}{z-w(z-1)}.$$
The residue then leaves
$$- m  \; \underset{z}{\mathrm{res}} \; 
\frac{1}{z^{m}} \log\frac{1}{1-z}
= -m \frac{1}{m-1}.$$
On flipping the sign we get $m/(m-1)$ which is the first term so we are
on the right track. Note that when $m=1$ this term will produce zero. For the residue at $w=z/(1-z)$ we write
$$- m  \; \underset{z}{\mathrm{res}} \; 
\frac{1}{z^{m}} \frac{1}{z-1} \log\frac{1}{1-z}
\; \underset{w}{\mathrm{res}} \;  
\frac{1}{w^{n+1}} \frac{1}{1-w} 
\frac{1}{w-z/(z-1)}.$$
Doing the evaluation of the residue yields
$$- m  \; \underset{z}{\mathrm{res}} \; 
\frac{1}{z^{m}} \frac{1}{z-1} \log\frac{1}{1-z}
\frac{(z-1)^{n+1}}{z^{n+1}} \frac{1}{1-z/(z-1)}
\\ = m  \; \underset{z}{\mathrm{res}} \; 
\frac{1}{z^{m+n+1}} \log\frac{1}{1-z}
(z-1)^{n+1}
\\ = m [z^{m+n}] \log\frac{1}{1-z}
(z-1)^{n+1}.$$
Using the cited formula a second time we put $n := m+n$ and $k := m-1$
to get
$$m \frac{1}{m-1} {m+n\choose m-1}^{-1}.$$
On flipping the sign we get the second term as required and we have the
claim.
Remark. In the above we have $m\gt 1.$ We get for $m=1$
$$[z^{n+1}] \log\frac{1}{1-z}
(z-1)^{n+1}
= \; \underset{z}{\mathrm{res}} \; 
\frac{1}{z^{n+2}} \log\frac{1}{1-z}
(z-1)^{n+1}.$$
Now we put $z/(z-1) = v$ so that $z = v/(v-1)$ and $dz = -1/(v-1)^2 \; 
dv$ to get
$$- \; \underset{v}{\mathrm{res}} \; 
\frac{1}{v^{n+2}} \log\frac{1}{1-v/(v-1)}
(v-1) \frac{1}{(1-v)^2}
\\ = \; \underset{v}{\mathrm{res}} \; 
\frac{1}{v^{n+2}} \frac{1}{1-v} \log(1-v).$$
On flipping the sign we obtain
$$\; \underset{v}{\mathrm{res}} \; 
\frac{1}{v^{n+2}} \frac{1}{1-v} \log\frac{1}{1-v} = H_{n+1},$$
again  as claimed.  This particular  value follows  by inspection,  of
course.
A: $$S_{m,n}(x)=\sum_{k=0}^{n}{m+k\choose m}^{-1} x^k=\Gamma(m+1)\sum_{k=0}^{n}\frac{ \Gamma (k+1)}{\Gamma (k+m+1)}x^k$$
$$S_{m,n}(x)=\, _2F_1(1,1;m+1;x)-\frac{\Gamma (m+1)\Gamma (n+2) }{\Gamma (m+n+2)} x^{n+1}\,
   _2F_1(1,n+2;m+n+2;x) $$
For $m \geq 2$
$$\, _2F_1(1,1;m+1;1)=\frac m{m-1}$$
Enjoy the simplification of
$$\frac{\Gamma (m+1)\Gamma (n+2) }{\Gamma (m+n+2)} \,
   _2F_1(1,n+2;m+n+2;1) $$
A: Choose $x_1, x_2, \dots, x_{m+n}$ uniformly at random from $[0,1]$. Let $A_k$ be the event that in the set $\{x_1, x_2, \dots, x_{m+k}\}$, the subset $\{x_m, \dots, x_{m+k-1}\}$ has the $k$ highest values, but the subset $\{x_m, \dots, x_{m+k}\}$ does not have the $k+1$ highest values.
(We consider $k=0, 1, \dots, n+1$; $A_0$ only checks the second condition, since the first is vacuous, and $A_{n+1}$ only checks the first condition, since the second is vacuous.)
Then the probability of $A_k$ is exactly $\binom{m+k}{m}^{-1} \cdot \frac{m-1}{m}$ for all $k \le n$, while the probability of $A_{n+1}$ is just $\binom{m+n}{n-1}^{-1}$. So the statement we want to show is equivalent to
$$
   \sum_{k=0}^n \frac{m}{m-1}\Pr[A_k] = \frac{m}{m-1}(1 - \Pr[A_{n+1}]) \iff \sum_{k=0}^{n+1} \Pr[A_k] = 1.
$$
To prove this, we show that exactly one of the events must happen:

*

*If $A_k$ happens, then one of the elements $x_1, \dots, x_{m-1}$ beats $x_{m+k}$, which prevents $\{x_m,\dots, x_{m+k'-1}\}$ from being the highest in $\{x_1, \dots, x_{m+k'}\}$ for any $k'>k$; then no $A_{k'}$ after $A_k$ can happen.

*Suppose $A_0$ does not happen; then $x_m$ is highest among $\{x_1, \dots, x_m\}$. Choose the first $k$ such that $x_{m+k}$ is less than $\max\{x_1, \dots, x_{m-1}\}$; then $A_k$ happens. Or, if there is no such $k$, then $A_{n+1}$ happens.

