Summation of powers of r and reciprocal of binomial coefficient

Question

Evaluate the following sum: $$\sum_{r=1}^m\frac{(m+1)(r-1)(m^{r-1})}{r\binom{m}{r}}$$

where $\binom{m}{r}$ stands for ${}^mC_r$

I initially tried to change this into $$\frac{m+1}{m}\sum_{r=1}^m\frac{m^r}{\binom{m}{r}}-\frac{m^r}{r\binom{m}{r}}$$ $$\Rightarrow \frac{m+1}{m} \left(\sum_{r=1}^m \frac{m^r}{\binom{m}{r}}-\frac{m^{r-1}}{\binom{m-1}{r-1}}\right)$$ to bring it into the form of $f(r)-f(r-1)$ type but this obviously got stuck, I had even tried bringing it in the form $$\frac{m+1}{m}\sum_{r=0}^{m-1}\frac{r.m^r}{\binom{m-1}{r}}$$ but it it got stuck again, could anyone please tell me how to proceed with my method or give me a better one? Thanks

Answer 1

We can use a telescoping approach.

We obtain \begin{align*} \color{blue}{\sum_{r=1}^m}&\color{blue}{\frac{(m+1)(r-1)m^{r-1}}{r\binom{m}{r}}}\\ &=\sum_{r=1}^m\frac{(mr-m+r-1)m^{r-1}}{r\binom{m}{r}}\\ &=\sum_{r=1}^m\frac{m^r}{\binom{m}{r}}-\sum_{r=1}^m\frac{(m-r+1)m^{r-1}}{r\binom{m}{r}}\\ &=\sum_{r=1}^m\frac{m^r}{\binom{m}{r}}-\sum_{r=1}^m\frac{m^{r-1}}{\binom{m}{r-1}}\tag{1}\\ &=\sum_{r=1}^m\frac{m^r}{\binom{m}{r}}-\sum_{r=0}^{m-1}\frac{m^{r}}{\binom{m}{r}}\tag{2}\\ &=\frac{m^m}{\binom{m}{m}}-\frac{m^0}{\binom{m}{0}}\tag{3}\\ &\,\,\color{blue}{=m^m-1} \end{align*}

Comment:

• In (1) we use the binomial identity $\binom{p}{q-1}=\binom{p}{q}\frac{q}{p-q+1}$.

• In (2) we shift the index of the right hand sum to start with $0$.

• In (3) we can cancel terms since the sums telescope.