Now, I was solving a this problem. It asks for summation of $$\sum\limits_{k =0}^\infty\dfrac{1}{{n+k \choose n}}$$ I solved it using this answer, the answer turns out to be $$\dfrac{n}{n-1}$$ However, can someone provide an explanation of how to go about proving this?
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Here is a calculation using a different integral for the beta function, also very simple. Suppose we seek to evaluate $$\sum_{n\ge 0} {n+q\choose q}^{-1}.$$
This is $$\sum_{n\ge 0} \frac{q! \times n!}{(n+q)!} = \sum_{n\ge 0} \frac{\Gamma(q+1) \times \Gamma(n+1)}{\Gamma(n+q+1)} \\ = \sum_{n\ge 0} (n+q+1) \frac{\Gamma(q+1) \times \Gamma(n+1)}{\Gamma(n+q+2)} = \sum_{n\ge 0} (n+q+1) \mathrm{B}(q+1, n+1).$$
Recall the beta function integral $$\mathrm{B}(x,y) = \int_0^\infty \frac{t^{x-1}}{(1+t)^{x+y}} dt.$$
This gives for the sum the representation $$\int_0^\infty \sum_{n\ge 0} (n+q+1) \frac{t^{q}}{(1+t)^{n+q+2}} dt = \int_0^\infty \frac{t^q}{(1+t)^{q+2}} \sum_{n\ge 0} (n+q+1) \frac{1}{(1+t)^n} dt \\ = \int_0^\infty \frac{t^q}{(1+t)^{q+2}} \frac{1+t}{t^2} dt + (q+1) \int_0^\infty \frac{t^q}{(1+t)^{q+2}} \frac{1+t}{t} dt \\ = \int_0^\infty \frac{t^{q-2}}{(1+t)^{q+1}} dt + (q+1) \int_0^\infty \frac{t^{q-1}}{(1+t)^{q+1}} dt.$$
Converting back from the beta functions that have appeared we get $$\mathrm{B}(q-1, 2) + (q+1) \mathrm{B}(q, 1) = \frac{\Gamma(q-1)\Gamma(2)}{\Gamma(q+1)} + (q+1) \frac{\Gamma(q)\Gamma(1)}{\Gamma(q+1)} \\ = \frac{1}{q(q-1)} + (q+1)\frac{1}{q} = \frac{q}{q-1}.$$
\begin{align}\sum^{\infty}_{k=0}\frac{1}{n+k\choose n}=\sum^{\infty}_{k=0}\frac{k!\cdot n!}{(n+k)!}&=\sum^{\infty}_{k=0}\frac{\Gamma{(n+1)}\cdot \Gamma{(k+1)}}{\Gamma(n+k+2)}\cdot (n+k+1)\\&=\sum^{\infty}_{k=0}(n+k+1)\cdot B(n+1,k+1)\end{align} Where $B(x,y)$ is the Beta function defined as $$B(x,y)=\int^{1}_{0}u^{x-1}(1-u)^{y-1}\,du$$ where $x,y>0$. So we can rewrite the last result as \begin{align}\sum^{\infty}_{k=0}(n+k+1)\cdot B(n+1,k+1)&=(n+1)\sum^{\infty}_{k=0}B(n+1,k+1)+\sum^{\infty}_{k=0}kB(n+1,k+1)\\ &=(n+1)\sum^{\infty}_{k=0}\int^{1}_{0}u^{k}(1-u)^{n}\,du+\sum^{\infty}_{k=0}k\int^{1}_{0}u^{k}(1-u)^{n}\,du\\ &=(n+1)\int^{1}_{0}(\sum^{\infty}_{k=0}u^{k})(1-u)^{n}\,du+\int^{1}_{0}u(\sum^{\infty}_{k=1}ku^{k-1})(1-u)^{n}\,du\\& =(n+1)\int^{1}_{0}\frac{1}{1-u}(1-u)^{n}\,du+\int^{1}_{0}u\frac{1}{(1-u)^2}(1-u)^{n}\,du\\& =(n+1)\frac{1}{n}+B(2,n-1)\\& =\frac{n+1}{n}+\frac{1}{n(n-1)}\\& =\frac{(n+1)(n-1)+1}{n(n-1)}\\& =\frac{n^2-1+1}{n(n-1)}=\frac{n}{n-1}\end{align}
Let U(n,k) be the term of your series.
You can prove that $U(n,k)=\frac{1}{(1+\frac{n}{1})...(1+\frac{n}{k})}$
you have: $U(n,k) = U(n,k-1)*\frac{k}{n+k} $
Then let $V(n,k) = k*U(n,k)$
You can prove that : $n*U(n,k) -U(n,k-1) = V(n,k) - V(n,k-1)$
Let $S(n,N)$ be the partial sum of the $U(n,k)$:
$S(n,N) = \sum_{k=0}^N U(n,k) = \sum\limits_{k =0}^N\dfrac{1}{{n+k \choose n}} $
By summing the relation (on the index k, from 1 to N) above you get:
$(n-1)*S(n,N) = n + V(n,N) - U(n,N)$
$U(n,N) \rightarrow 0$ ; $V(n,N) \rightarrow 0$ , when $ N \rightarrow \infty$
So you get: $L = \frac{n}{n-1} $
You could evaluate it using the Gauss hypergeometric Function $_2F_1$. We have: $$ _2F_1(a,b,c;z)=\sum_{k=0}^{\infty}\frac{\Gamma(a+k)\Gamma(b+k)\Gamma({c})}{\Gamma(a)\Gamma(b)\Gamma(c+k)\Gamma(k+1)}\cdot z^k $$ And: $$ _2F_1(a,b,c;1)=\sum_{k=0}^{\infty}\frac{\Gamma(a+k)\Gamma(b+k)\Gamma({c})}{\Gamma(a)\Gamma(b)\Gamma(c+k)\Gamma(k+1)}=\frac{\Gamma({c})\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)} $$ Rewriting your sum yields: $$ \sum_{k=0}^{\infty} \frac{1}{\binom{n+k}{n}}=\sum_{k=0}^{\infty} \frac{n!\cdot k!}{(n+k)!}=\sum_{k=0}^{\infty} \frac{\Gamma(n+1)\Gamma(k+1)}{\Gamma(n+k+1)}=\sum_{k=0}^{\infty} \frac{n!\cdot k!}{(n+k)!}=\sum_{k=0}^{\infty} \frac{\Gamma(k+1)\Gamma(k+1)\Gamma(n+1)}{\Gamma(1)\Gamma(1)\Gamma(n+k+1)\Gamma(k+1)}=_2F_1(1,1,n+1;1)=\frac{\Gamma({n+1})\Gamma(n+1-1-1)}{\Gamma(n+1-1)\Gamma(n+1-1)}=\frac{\Gamma({n+1})\Gamma(n-1)}{\Gamma(n)\Gamma(n)}=\frac{n\Gamma({n})\Gamma(n-1)}{\Gamma(n)\cdot n\Gamma(n-1)}=\frac{n}{n-1} $$
