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?


  • $\begingroup$ What range of $r$ are you interested in? $\endgroup$ May 19 '14 at 18:51

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.