# Evaluate $\sum_{r=1}^{m} \frac{(r-1)m^{r-1}}{r\cdot\binom{m}{r}}$

Evaluate:$$\sum_{r=1}^{m} \frac{(r-1)m^{r-1}}{r\cdot\binom{m}{r}}$$

Using the property:$$r\binom{m}{r}=m\binom{m-1}{r-1}$$

It is same as $$\sum_{r=2}^{m} \frac{(r-1)m^{r-1}}{m\cdot\binom{m-1}{r-1}}$$

How I do now?

• Right from the beginning we have $\sum\limits_{r=\color{red}1}^{m} \frac{(r-1)m^{r-1}}{r\cdot\binom{m}{r}}=\sum\limits_{r=\color{red}2}^{m} \frac{(r-1)m^{r-1}}{r\cdot\binom{m}{r}}$ Commented Jul 31, 2020 at 14:23
• @callculus yes you are right Commented Jul 31, 2020 at 14:46

Let $$S=\sum_{r=1}^{m} \frac{(r-1)m^{r-1}}{r\cdot\binom{m}{r}}$$

Multiply both sides by $$m+1$$.

$$S(m+1)=\sum_{r=1}^{m} \frac{(m+1)(r-1)m^{r-1}}{r\cdot\binom{m}{r}}$$

$$S(m+1)=\sum_{r=1}^{m} \frac{(mr-(m-r+1))m^{r-1}}{r\cdot\binom{m}{r}}$$

$$S(m+1)=\sum_{r=1}^{m} \frac{(rm^r-(m-r+1)m^{r-1})}{r\cdot\binom{m}{r}}$$

$$S(m+1)=\sum_{r=1}^{m} \frac{m^r}{\binom{m}{r}}-\frac{(m-r+1)m^{r-1}r!\cdot(m-r)!} {r\cdot m!}$$

$$S(m+1)=\sum_{r=1}^{m} \frac{m^r}{\binom{m}{r}}-\frac{m^{r-1}(m-r+1)!(r-1)!}{m!}$$

$$S(m+1)=\sum_{r=1}^{m} \frac{m^r}{\binom{m}{r}}-\frac{m^{r-1}}{\binom{m}{r-1}}$$

Now this becomes a telescoping series.

$$\boxed {S(m+1)=m^m-1}$$

• Does my comment answer your question? Or have I misunderstood your question? Commented Jul 31, 2020 at 14:36
• @callculus yes you were right,how did you do then? Commented Jul 31, 2020 at 14:49
• Then I´ve done nothing. For me it looks o.k. what you´ve done. Commented Jul 31, 2020 at 15:12