Another sum involving binomial coefficients.

Question

Let both $a$,$b$ and $\theta$ be real numbers not equal to a negative integer. Let $m$ be a positive integer. I have shown that the following equality holds: \begin{eqnarray} &&S^{a}_{b,\theta} (m) := \sum\limits_{i=0}^{m-1} \binom{i+a}{b} \frac{1}{i+\theta} = \\ &&\sum\limits_{l=1}^\infty \binom{a-\theta}{b-l} \frac{1}{l} \left(\binom{\theta-1+m}{l} - \binom{\theta-1}{l}\right) + \\&&\binom{a-\theta}{b} \left(\psi(\theta+m) - \psi(\theta)\right) =\\ &&\binom{a-\theta}{b-1} \left[ (\theta+m-1) F_{4,3}\left[\begin{array}{rrrr} 1 & 1 & 1-b & 2-m-\theta \\ 2 & 2 & 2+a-b-\theta &\end{array};1\right] - (\theta-1) F_{4,3}\left[\begin{array}{rrrr} 1 & 1 & 1-b & 2-\theta \\ 2 & 2 & 2+a-b-\theta &\end{array};1\right] \right] +\\ &&\binom{a-\theta}{b} \left(\psi(\theta+m) - \psi(\theta)\right) \end{eqnarray} Here $\psi$ is the polygamma function of order zero. Now there is a question. What happens if the binomial factor in the sum is replaced by a product of two different binomial factors each one with different values of $a$ and $b$.

Answer 1

We prove the identity for $b=n$ being a positive integer. We have:

Take $n=1$: \begin{equation} S^a_{1,\theta}(m) = \sum\limits_{i=0}^{m-1} \left(1 + \frac{a-\theta}{i+\theta}\right) = m + \binom{a-\theta}{1} \left(\psi(m+\theta) - \psi(\theta)\right) \end{equation}

Now, take $n=2$: \begin{equation} S^a_{2,\theta}(m) = \sum\limits_{i=0}^{m-1}\left( \frac{1}{2}(-1+2 a +i - \theta) + \frac{\binom{a-\theta}{2}}{i+\theta} \right)= \frac{1}{4} m(-3+ 4 a + m - 2 \theta) + \binom{a-\theta}{2} \left(\psi(m+\theta) - \psi(\theta)\right) \end{equation}

For generic $n$ we split our sum into following two terms: \begin{equation} S^a_{n,\theta}(m) = \sum\limits_{i=0}^{m-1} \frac{\binom{i+a}{n} - \binom{a-\theta}{n}}{i+\theta} + \binom{a-\theta}{n} \left(\psi(m+\theta) - \psi(\theta)\right) \end{equation} It is readily seen that the term in the sum on the right hand side is always a polynomial in $i+\theta$ and thus that sum must be a polynomial in $m$. Now the only thing we need to do is to prove that: \begin{equation} \sum\limits_{i=0}^{m-1} \frac{\binom{i+a}{n} - \binom{a-\theta}{n}}{i+\theta} \stackrel{?}{=} \sum\limits_{l=1}^n \binom{a-\theta}{n-l} \frac{1}{l} \left(\binom{\theta-1+m}{l} - \binom{\theta-1}{l}\right) \end{equation} Let us differentiate both sides with respect to $m$. We have: \begin{equation} lhs = \frac{\binom{m-1+a}{n} - \binom{a-\theta}{n}}{m-1+\theta} \end{equation} and \begin{equation} rhs = \sum\limits_{l=1}^n \binom{a-\theta}{n-l} \frac{1}{l} \left(\binom{\theta-1+m}{l} - \binom{\theta-2+m}{l} \right) = \sum\limits_{l=1}^n \binom{a-\theta}{n-l} \frac{1}{l} \binom{\theta-2+m}{l-1} = \frac{1}{\theta-1+m}\sum\limits_{l=1}^n \binom{a-\theta}{n-l} \binom{\theta-1+m}{l} =\frac{\binom{a-1+m}{n}-\binom{a-\theta}{n}}{\theta-1+m} \end{equation} where in the last equality we used the Chu-Vandermonde identity. This completes the proof.

Answer 2

By using differentiation and integration of power functions we convert the sum in question to a geometric series. After summing up that series we obtain the following formula. \begin{equation} \left.S^a_{b,\theta}(m) = \int\limits_0^1 \xi^{\theta-1} \frac{1}{b!} \frac{d^b}{d x^b} \left(\frac{x^a - x^{a+m} \xi^m}{1 - x \xi}\right) d \xi \right|_{x=1} \end{equation} Now, we use the chain rule to calculate the derivative. We have: \begin{eqnarray} S^a_{b,\theta}(m) = \int\limits_0^1 \xi^{\theta-1} \frac{1}{b!} \sum\limits_{j=0}^n \binom{b}{j} \left(a_{(b-j)} x^{a-b+j} - (a+m)_{(b-j)} x^{a+m-b+j} \xi^m\right) \cdot \frac{\xi^j j!}{(1 - x \xi)^{j+1}} d\xi \end{eqnarray} Simplifying the result we have: \begin{eqnarray} \left.S^a_{b,\theta}(m) = \sum\limits_{j=0}^b \left[ \binom{a}{b-j} \int\limits_0^1 \frac{\xi^{\theta+j-1}}{(1 - x \xi)^{j+1}} d\xi - \binom{a+m}{b-j} \int\limits_0^1 \frac{\xi^{\theta+m+j-1}}{(1 - x \xi)^{j+1}} d\xi \right] \right|_{x=1} \end{eqnarray} Note that the integrals at $x=1$ are divergent and as such the expression is not defined. Therefore we have to proceed with caution. We do take the limit $x=1$ but we introduce a small $\epsilon$ to avoid obtaining infinities. We have: \begin{eqnarray} S^a_{b,\theta}(m) &=& \sum\limits_{j=0}^b \left[ \binom{a}{b-j} B(\theta+j,-j+\epsilon) - \binom{a+m}{b-j} B(\theta+j+m,-j+\epsilon) \right] \\ &=& \frac{\pi}{\sin(\pi \epsilon)}\sum\limits_{j=0}^b \left[ \binom{a}{b-j}\frac{(\theta+j-1)!}{(j-\epsilon)!(\theta-1+\epsilon)!} - \binom{a+m}{b-j} \frac{(\theta+m+j-1)!}{(j-\epsilon)!(\theta+m-1+\epsilon)!} \right] \\ \end{eqnarray} where in the second line we used the definition of the beta function and then we used the reflection formula for the Gamma function. Now we have to be careful in taking the limit $\epsilon \rightarrow 0$. Since: \begin{eqnarray} \sum\limits_{j=0}^b (-1)^j \binom{a}{b-j} \binom{\theta+j-1}{j} = \sum\limits_{j=0}^b (-1)^j \binom{a+m}{b-j} \binom{\theta+m+j-1}{j} = \binom{a-\theta}{b} \end{eqnarray} we can add and subtract certain terms inside the square bracket. We have: \begin{eqnarray} S^a_{b,\theta}(m) &=& \frac{\pi \epsilon}{\sin(\pi \epsilon)} \sum\limits_{j=0}^b (-1)^j \left[ \binom{a}{b-j} (\theta+j-1)! \left(\frac{\frac{1}{(j-\epsilon)!(\theta-1+\epsilon)!}-\frac{1}{(j)!(\theta-1)!}}{\epsilon}\right) - \binom{a+m}{b-j} (\theta+m+j-1)! \left(\frac{\frac{1}{(j-\epsilon)!(\theta+m-1+\epsilon)!}-\frac{1}{(j)!(\theta+m-1)!}}{\epsilon}\right) \right] \end{eqnarray} Now we can take the limit term by term and we readily arrive at the following formula: \begin{eqnarray} S^a_{b,\theta}(m) &=& \sum\limits_{j=0}^b (-1)^j \psi(1+j) \left[ \binom{a}{b-j} \binom{\theta+j-1}{j} - \binom{a+m}{b-j} \binom{\theta+m+j-1}{j} \right] + \binom{a-\theta}{b} \left(\psi(\theta+m) - \psi(\theta)\right) \\ &=& \sum\limits_{j=0}^b H_j \left[ \binom{a}{b-j} \binom{-\theta}{j} - \binom{a+m}{b-j} \binom{-\theta-m}{j} \right] + \binom{a-\theta}{b} \left(\psi(\theta+m) - \psi(\theta)\right) \end{eqnarray} Here $H_j$ is the harmonic number.

Answer 3

Here we are going to do the sum in yet another way. The purpose of this is to show that the sum can be done using elementary methods. Firstly let us assume that $\theta= a$. Then since \begin{equation} \binom{i+a}{b} \frac{1}{i + a} = \frac{1}{b} \binom{i+a-1}{b-1} \end{equation} we readily have: \begin{equation} S^a_{b,a} = \frac{1}{b} \left[ \binom{a-1+m}{b} - \binom{a-1}{b} \right] \end{equation} Now take $\theta = a-1$. It is also not hard to see that we have: \begin{equation} \binom{i+a}{b} \frac{1}{i + a-1} = \frac{1}{b} \binom{i+a-1}{b-1} + \frac{1}{b(b-1)} \binom{i+a-2}{b-2} \end{equation} Thus we readily have: \begin{equation} S^a_{b,a-1} = \frac{1}{b} \left[ \binom{a-1+m}{b} - \binom{a-1}{b} \right] + \frac{1}{b(b-1)} \left[ \binom{a-2+m}{b-1} - \binom{a-2}{b-1} \right] \end{equation} Now, in general, let us assume that $n := a- \theta$ is a positive integer. Then we have: \begin{eqnarray} \binom{i+a}{b} \frac{1}{i + a-n} &=& (i+a)_{(n)} \frac{(i+a-n-1)!}{b!(i+a-b)!}\\ &=& \sum\limits_{l=1}^{n+1} n_{(l-1)} (i+a-l)_{(n+1-l)} \frac{(i+a-n-1)!}{b!(i+a-b)!}\\ &=& \sum\limits_{l=1}^{n+1} \frac{n_{(l-1)}}{b_{(l)}} \binom{i+a-l}{b-l} \end{eqnarray} Now the sum over $i$ is straightforward . We have: \begin{equation} S^a_{b,a-n} = \sum\limits_{l=1}^{n+1} \frac{n_{(l-1)}}{b_{(l)}} \left[ \binom{a-l+m}{b-l+1} - \binom{a-l}{b-l+1}\right] \end{equation} Simplifying the above we get: \begin{eqnarray} S^a_{b,\theta} = \sum\limits_{l=1}^{a-\theta+1} \frac{\binom{a-\theta}{l-1}}{\binom{b}{l}} \frac{1}{l} \left[ \binom{a-l+m}{b-l+1} - \binom{a-l}{b-l+1}\right] \end{eqnarray} The only question that remains is what happens if $a-\theta$ is a real number..