Prove $B(p,q) + B(p+1,q) + B(p+2,q) + ... = B(p,q-1) $ where $B$ is beta function and $q > 1$ Prove $B(p,q) + B(p+1,q) + B(p+2,q) + ... = B(p,q-1) $ where $B$ is beta function and $q &gt; 1$.

I tried with some basic formulas for beta function, mathematical induction but just cannot get some good idea and solve this. If anyone have some idea, please help. I will be grateful.
$$B(p,q) + B(p+1,q) + B(p+2,q) + ...=\int_0^1 t^{p-1}(1-t)^{q-1}dt+\int_0^1 t^{p}(1-t)^{q-1}dt+\int_0^1 t^{p+1}(1-t)^{q-1}dt+...$$
$$=\int_0^1 (t^{p-1}(1-t)^{q-1}+t^{p}(1-t)^{q-1}+t^{p+1}(1-t)^{q-1}+...)dt$$
$$=\int_0^1 \frac{t^{p-1}(1-t)^{q-1}}{1-t} dt$$
$$=\int_0^1 t^{p-1}(1-t)^{q-2} dt$$
$$=B(p,q-1)$$
By definition, starting from the LHS of your problem,
$$
\sum_{n=p}^\infty\beta(n,q) = \sum_{n=p}^\infty\int_0^1t^{n-1}(1-t)^{q-1}dt 
$$
Since $t$ is always in $[0,1]$ things converge nicely (I think uniformly) and we are free to interchange integral and sum:
$$
\sum_{n=p}^\infty\beta(n,q) = \int_0^1\sum_{n=p}^\infty t^{n-1}(1-t)^{q-1}dt 
$$
Again, for this range of $t$, the sums converge:
$$
\sum_{n=p}^\infty t^{n-1}(1-t)^{q-1} = t^{p-1}\sum_{m=0}^\infty t^{m}(1-t)^{q-1}
= t^{p-1}\frac1{1-t}(1-t)^{q-1}=  t^{p-1}(1-t)^{q-2}
$$
Finally, placing this back into the integral, we have
$$
\sum_{n=p}^\infty\beta(n,q) =  \int_0^1t^{p-1}(1-t)^{q-2}dt =\beta(p,q-1)
$$