Show that $\frac{1}{r!}-\frac{1}{(r+1)!}\equiv\frac{r}{(r+1)!}$.

Question

Show that $\frac{1}{r!}-\frac{1}{(r+1)!}\equiv\frac{r}{(r+1)!}$.

I get $$\frac{1}{r!}-\frac{1}{(r+1)!}=\frac{(r+1)-r!}{r!(r+1)!}$$

and in the numerator since $$(r+1)!-r!=r$$

so $$\frac{(r+1)-r!}{r!(r+1)!}=\frac{r}{r!(r+1)!}$$

How do I deal with the extra $r!$ in the denominator?

Answer 1

$$\frac1{r!}=\frac{(r+1)}{(r+1)\cdot r!}=\frac{(r+1)}{(r+1)!}$$

So, $$\frac1{r!}-\frac1{(r+1)!}=\frac{r+1-1}{(r+1)!}=\frac r{(r+1)!} $$

Where have you found $r!$ as the multiplier of the denominator?

Answer 2

Your first subtraction is incorrect: You want

$$\frac{1}{r!} - \frac{1}{(r + 1)!} = \frac{r + 1}{(r + 1)!} - \frac{1}{(r + 1)!}$$

Answer 3

Multiplying the first term by $(r+1)/(r+1)$ gives

$$\frac{1}{r!}-\frac{1}{(r+1)!}=\frac{r+1-1}{(r+1)!}=\frac{r}{(r+1)!}$$

as desired. In your argument, the equality $(r+1)!-r!=r$ only holds for $r=1$ (assuming $r$ is a positive integer). Also, there are some factorials missing in some $(r+1)$ terms.