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How to prove that $$ S = \sum_{k=0}^{K-1} \binom{k+K-1}{K-1} \frac{a^k b^K + a^K b^k}{(a+b)^{k+K}} \mathop{=}\limits^{?} 1, $$ For $K=2$, I have S = $\frac{a^2+b^2}{(a+b)^2} + \frac{2ab(a+b)}{(a+b)^3}$, which can be boiled down as $S = \frac{(a+b)(a^2+b^2+2ab)}{(a+b)^3} = 1$.

Also, WolframAlpha gave the result $$ S = 2 - \binom{2K-1}{K-1} \frac{(xy)^K}{(x+y)^{2K}} \left\{ {}_2F_1\left(1,2K;K+1;\frac{a}{a+b} \right) + {}_2F_1\left(1,2K;K+1;\frac{b}{a+b} \right) \right\}, $$ but $S = 1$ only when $0 < \frac{a}{a+b} < 1$ and $0 < \frac{b}{a+b} < 1$.

Kyle
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3 Answers3

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Write $x={a\over a+b},\ J=K-1$, and $$f(x,J) = (1-x)^{J+1}\sum_{k=0}^J{J+k\choose k}x^k$$ Then we want to show $$f(x,J)+f(1-x,J)=1,\ J=0,1,2,\dots\tag{1}$$

We proceed by induction on $J$. When $J=0$, $(1)$ says $x+(1-x)=1.$

Suppose that $J>0$ and that $(1)$ is true for $J-1$. Then $$\begin{align} f(x,J)&=(1-x)^{J+1}\sum_{k=0}^J{J-1+k\choose k}x^k+(1-x)^{J+1}\sum_{k=0}^J{J-1+k\choose k-1}x^k\tag{2} \end{align}$$ The first term on the right of $(2)$ is $$ (1-x)^{J+1}\sum_{k=0}^{J-1}{J-1+k\choose k}x^k+(1-x)^{J+1}{2J-1\choose J}x^J$$ or $$(1-x)f(x,J-1)+(1-x)^{J+1}x^J{2J-1\choose J}\tag{3}$$ The second term on the right of $(2)$ is $$\begin{align} (1-x)^{J+1}\sum_{k=1}^J {J-1+k\choose k-1}x^k&=(1-x)^{J+1}\sum_{k=0}^{J-1}{J+k\choose k}x^{k+1}\\ &=xf(x,J)-(1-x)^{J+1}{2J\choose J}x^{J+1}\tag{4} \end{align}$$ Now $(2),(3),\text{ and }(4)$ give

$ (1-x)f(x,J)=(1-x)f(x,J-1)+(1-x)^{J+1}x^J{2J-1\choose J}-(1-x)^{J+1}{2J\choose J}x^{J+1}\tag{6} $

By symmetry,

$ xf(1-x,J)=xf(1-x,J-1)+x^{J+1}(1-x)^J{2J-1\choose J}-x^{J+1}{2J\choose J}(1-x)^{J+1}\tag{7} $

Multiply $(6)$ by $x$, multiply $(7)$ by $1-x$, add the results and apply the induction hypothesis to get $$ x(1-x)(f(x,J)+f(1-x,J))= x(1-x)\tag{8}$$ after verifying that $${2J\choose J}=2{2J-1\choose J}$$

This proves $(2)$ for $x\neq0,1$, but since it is a polynomial identity, it is true for these values also.

saulspatz
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  • Just a brief query, do you understand the downvote I received? – Marko Riedel May 30 '19 at 15:27
  • @MarkoRiedel No, I was wondering about that that myself. I haven't upvoted it yet, because I haven't finished reading it. (I'm still looking for typos in my answer, so I've only had a chance to skim yours.) – saulspatz May 30 '19 at 15:28
  • Thank you for this answer, the idea of using induction hypothesis is so simple, I upvoted this post anyway. – Kyle May 31 '19 at 04:46
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Starting from

$$S = \sum_{q=0}^{K-1} {K-1+q\choose K-1} \frac{a^q b^K + a^K b^q}{(a+b)^{q+K}}$$

we get two pieces

$$\frac{b^K}{(a+b)^K} \sum_{q=0}^{K-1} {K-1+q\choose K-1} \frac{a^q}{(a+b)^q} \\ + \frac{a^K}{(a+b)^K} \sum_{q=0}^{K-1} {K-1+q\choose K-1} \frac{b^q}{(a+b)^q}.$$

This is

$$\frac{b^K}{(a+b)^K} [z^{K-1}] \frac{1}{1-z} \frac{1}{(1-az/(a+b))^K} \\ + \frac{a^K}{(a+b)^K} [z^{K-1}] \frac{1}{1-z} \frac{1}{(1-bz/(a+b))^K}.$$

Call these $S_1$ and $S_2.$ The first sum is

$$S_1 = \frac{b^K}{(a+b)^K} \mathrm{Res}_{z=0} \frac{1}{z^K} \frac{1}{1-z} \frac{1}{(1-az/(a+b))^K} \\ = b^K \mathrm{Res}_{z=0} \frac{1}{z^K} \frac{1}{1-z} \frac{1}{(a+b-az)^K} \\ = \frac{b^K}{a^K} \mathrm{Res}_{z=0} \frac{1}{z^K} \frac{1}{1-z} \frac{1}{((a+b)/a-z)^K} \\ = (-1)^{K+1} \frac{b^K}{a^K} \mathrm{Res}_{z=0} \frac{1}{z^K} \frac{1}{z-1} \frac{1}{(z-(a+b)/a)^K}.$$

Now residues sum to zero so we compute this from the residues at the poles at $z=1$ and $z=(a+b)/a.$ The residue at infinity is zero by inspection. The residue at $z=1$ is

$$(-1)^{K+1} \frac{b^K}{a^K} \frac{1}{(1-(a+b)/a)^K} = (-1)^{K+1} b^K \frac{1}{(a-(a+b))^K} \\ = (-1)^{K+1} b^K \frac{1}{(-b)^K} = -1.$$

For the residue at $z=(a+b)/a$ we require

$$\frac{1}{(K-1)!} \left(\frac{1}{z^K} \frac{1}{z-1} \right)^{(K-1)} \\ = \frac{1}{(K-1)!} \sum_{q=0}^{K-1} {K-1\choose q} (-1)^q \frac{(K-1+q)!}{(K-1)!} \frac{1}{z^{K+q}} (-1)^{K-1-q} \frac{(K-1-q)!}{(z-1)^{K-q}} \\ = (-1)^{K+1} \sum_{q=0}^{K-1} {K-1+q\choose K-1} \frac{1}{z^{K+q}} \frac{1}{(z-1)^{K-q}}.$$

Evaluating the residue we find

$$\left. (-1)^{K+1} \frac{b^K}{a^K} (-1)^{K+1} \sum_{q=0}^{K-1} {K-1+q\choose K-1} \frac{1}{z^{K+q}} \frac{1}{(z-1)^{K-q}} \right|_{z=(a+b)/a} \\ = \frac{b^K}{a^K} \sum_{q=0}^{K-1} {K-1+q\choose K-1} \frac{a^{K+q}}{(a+b)^{K+q}} \frac{1}{((a+b)/a-1)^{K-q}} \\ = \sum_{q=0}^{K-1} {K-1+q\choose K-1} \frac{a^{K+q}}{(a+b)^{K+q}} \frac{b^q}{a^q} \frac{b^{K-q}}{a^{K-q}}\frac{1}{((a+b)/a-1)^{K-q}} \\ = \sum_{q=0}^{K-1} {K-1+q\choose K-1} \frac{a^{K+q}}{(a+b)^{K+q}} \frac{b^q}{a^q} \\ = \frac{a^K}{(a+b)^K} \sum_{q=0}^{K-1} {K-1+q\choose K-1} \frac{b^q}{(a+b)^{q}} = S_2.$$

We recognise $S_2$ and hence we have shown that

$$S_1-1+S_2 = 0$$

or

$$\bbox[5px,border:2px solid #00A000]{ \sum_{q=0}^{K-1} {K-1+q\choose K-1} \frac{a^q b^K + a^K b^q}{(a+b)^{q+K}} = 1}$$

as claimed.

Addendum. This is a special case with $x=a/(a+b)$ of the identity at this MSE link.

Marko Riedel
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  • What theorem are you referring to when you say "residues sum to zero"? Can you give a link to proof? I remember that for a rational function ${p(z)\over q(z)}$ with $\deg q -\deg p\geq2$ the sum of the residues is $0$, but then you wouldn't be mentioning $\infty.$ (Of course the result I remember is sufficient in this case, and it's easy to prove.) – saulspatz May 30 '19 at 16:21
  • The rational example that you cite also includes the residue at infinity which is zero in your case. I simply use the generic version, where we have a circular contour going to infinity and eventually enclosing all isolated singularities of a holomorphic function. So you are using that residue as well, even though it is encapsulated in the zero value and not specified explicitly. There is a proof at this MSE link. – Marko Riedel May 30 '19 at 17:34
  • When we use the circular contour the contribution from the residue at infinity becomes a requirement, as can be seen at the following two links, MSE link I and MSE link II. Also consult MathWorld. – Marko Riedel May 30 '19 at 17:34
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It needs to be proven: $$S = \sum_{k=0}^{K-1} \binom{k+K-1}{K-1} \frac{a^k b^K + a^K b^k}{(a+b)^{k+K}} \mathop{=}\limits^{?} 1$$ For $K=2$: $$S = {1\choose 1}\frac{b^2+a^2}{(a+b)^2}+{2\choose 1}\frac{ab^2+a^2b}{(a+b)^3}=\\ \frac{1}{(a+b)^3}\left[{1\choose 1}(b^2+a^2)\color{blue}{(a+b)}+\color{green}{2\choose 1}(ab^2+a^2b)\right]=\\ \frac{1}{(a+b)^3}\left[a^3+\left[\color{blue}{1\choose 1}+\color{green}{2\choose 1}\right](a^2b+ab^2)+b^3\right]=1$$ Note: $1={1\choose 0}={1\choose 1}={3\choose 0}={3\choose 3}, {1\choose 1}+{2\choose 1}={3\choose 1}$.

For $K=3$: $$S = {2\choose 2}\frac{b^3+a^3}{(a+b)^3}+{3\choose 2}\frac{ab^3+a^3b}{(a+b)^4}+{4\choose 2}\frac{a^2b^3+a^3b^2}{(a+b)^5}=\\ \frac1{(a+b)^5}\left[{2\choose 2}(b^3+a^3)\color{blue}{(a+b)^2}+\color{red}{{3\choose 2}}(ab^3+a^3b)\color{green}{(a+b)}+\color{brown}{4\choose 2}(a^2b^3+a^3b^2)\right]=\\ \frac1{(a+b)^5}\left[a^5+b^5+\left[\color{blue}{{2\choose 1}}+\color{red}{{3\choose 2}}\color{green}{1\choose 0}\right](a^4b+ab^4)+\left[\color{blue}{2\choose 2}+\color{red}{3\choose 2}\color{green}{1\choose 1}+\color{brown}{4\choose 2}\right](a^3b^2+a^2b^3)\right]=1.$$ For $K=4$: $$S = {3\choose 3}\frac{b^4+a^4}{(a+b)^4}+{4\choose 3}\frac{ab^4+a^4b}{(a+b)^5}+{5\choose 4}\frac{a^2b^4+a^4b^2}{(a+b)^6}+{6\choose 3}\frac{a^3b^4+a^4b^3}{(a+b)^7}=\\ \frac1{(a+b)^7}\left[{3\choose 3}(b^4+a^4)\color{blue}{(a+b)^3}+\color{red}{{4\choose 3}}(ab^4+a^4b)\color{green}{(a+b)^2}+\color{purple}{{5\choose 3}}(a^2b^4+a^4b^2)\color{brown}{(a+b)}+\color{brown}{{6\choose 3}}(a^3b^4+a^4b^3)\right]=\\ \frac1{(a+b)^7}\left[a^7+b^7+\left[\color{blue}{{3\choose 1}}+\color{red}{{4\choose 3}}\color{green}{2\choose 0}\right](a^6b+ab^6)+\left[\color{blue}{{3\choose 2}}+\color{red}{{4\choose 3}}\color{green}{{2\choose 1}}+\color{purple}{{5\choose 3}}\right](a^5b^2+a^2b^5)+\left[\color{blue}{3\choose 0}+\color{red}{{4\choose 3}}\color{green}{2\choose 0}+\color{purple}{5\choose 3}\color{brown}{1\choose 0}+\color{brown}{{6\choose 3}}\right](a^4b^3+a^3b^4)\right]=1.$$ The sums of binomials should be calculated by closed form, which I will leave to you.

farruhota
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