# An inverse binomial summation.

I am looking for a closed form for this summation: $$\sum_{j=1}^m\frac{r^{-j}}{j{m\choose j}} = \sum_{j=1}^m\frac{r^{-j}}{m{m-1\choose j-1}} = \frac1{rm} \sum_{k=0}^{m-1}\frac{r^{-k}}{{m-1\choose k}}$$ I looked at some tables of binomial sums but I couldn't find anything similar to this. Could anyone help me to simplify this summation? Or find an upper bound?

Thanks.

• What range of $r$ are you interested in? – Greg Martin May 19 '14 at 18:51

## 1 Answer

The following doesn't give a closed form but does provide an alternative form for the sum that you may find helpful.

Assuming $r \neq 0$ and $r \neq -1$, one can generalize the result of this answer, which analyzes the case $r=1$, to get an equivalent sum. We have

$$\frac{1}{m}\sum_{k=0}^{m-1}\frac{r^{-k}}{{m-1 \choose k}} = \frac{1}{m}\sum_{k=0}^{m-1}\frac{\Gamma(k+1)\Gamma(m-k)}{\Gamma(m)}r^{-k} =$$ $$\sum_{k=0}^{m-1}\beta(k+1, m-k)r^{-k}$$

Where $\Gamma$ is the gamma function and $\beta$ is the beta function. Using the integral form of the beta function this sum equals

$$\sum_{k=0}^{m-1}r^{-k}\int_{0}^{1}t^k(1-t)^{m-k-1}\operatorname{d}t = \int_{0}^{1}\sum_{k=0}^{m-1}\left(\frac{t}{r}\right)^k(1-t)^{m-k-1} \operatorname{d} t =$$ $$\int_{0}^{1}\frac{\left(\frac{t}{r}\right)^m - (1-t)^m}{\frac{t}{r} - (1-t)} \operatorname{d} t$$

Making the substituion $u = \left(\frac{t}{r}\right) - (1-t)$ assuming $r \neq -1$ brings this to the form $$\frac{r}{\left(r+1\right)^{m+1}}\int_{-1}^{\frac{1}{r}}\frac{ (u+1)^{m+1} - (1-ru)^m}{u}\operatorname{d} u =$$

$$\frac{r}{(r+1)^{m+1}}\left[\int_{-1}^{\frac{1}{r}}\frac{(u+1)^m - 1}{u} \operatorname{d}u + r\int_{-1}^{1/r}\frac{1 - (1 - ru)^m}{ru} \operatorname{d}u \right] =$$

$$\frac{r}{(r+1)^{m+1}}\left[\int_{-1}^{\frac{1}{r}}\sum_{k=0}^{m-1}(1+u)^k \operatorname{d}u + r\int_{-1}^{\frac{1}{r}}\sum_{k=0}^{m-1}(1-ru)^k \operatorname{d} u\right] =$$

$$\frac{r}{\left(r+1\right)^{m+1}}\sum_{k=0}^{m-1}\left[\frac{\left(1 + \frac{1}{r}\right)^{k+1}}{k+1} + \frac{(1+r)^{k+1}}{k+1}\right] =$$

$$r\sum_{k=0}^{m-1}\frac{\left(1+r\right)^{k-m}}{k+1}\left(1 + \frac{1}{r^{k+1}} \right)$$

From whence it follows that

$$\sum_{j=1}^{m}\frac{r^{-j}}{j {m \choose j}} = \frac{1}{rm}\sum_{k=0}^{m-1} \frac{r^{-k}}{{m-1 \choose k}} =$$

$$\sum_{k=0}^{m-1}\frac{\left(1+r\right)^{k-m}}{k+1}\left(1 + \frac{1}{r^{k+1}} \right)$$

Feeding this new form of the sum to Wolfram Alpha gave a closed form in terms of special functions, but it doesn't seem to be valid for all $r \notin \left\{0,-1\right\}$. It doesn't seem likely that a nice closed form exists, but perhaps the new form I've shown you will help you find an upper bound that suits your purposes.

The case $r = -1$ must be handled separately. A closed form for this case is given in this paper, along with a more general form of the result I just proved.