combinatorial formula closed form

Question

I have the following sum and I am struggling to show that the closed form is given by the R.H.S of (1).

$$ \sum^\tau _{j=1} \frac{(1-x)^{j-1}x}{1-(1-x)^\tau}(1-(1-x)^j) = \frac{1-(1-x)^{\tau+1}}{2-x} \tag 1 $$ Any help is appreciated as I am quite lost!

Answer 1

Hint Denote $y:=1-x$. Then you need to show

$$\sum^\tau _{j=1} \frac{y^{j-1}(1-y)}{1-y^\tau}(1-y^j) = \frac{1-y^{\tau+1}}{1+y}$$

or equivalently $$\sum^\tau _{j=1} y^{j-1}(1-y^j) = \frac{1-y^{\tau+1}}{1+y}\frac{1-y^{\tau}}{1-y}$$

The LHS splits into two sums of geometric sequences, for which you have closed form.

Answer 2

Rewrite the LHS as $$ \frac{x}{1-(1-x)^\tau}\left(\sum_{j=1}^\tau (1-x)^{j-1}-\sum_{j=1}^\tau (1-x)^{2j-1}\right) $$ There are two summations above which are both finite geometric series, so you can find closed forms for both of them. Do so and simplify.

Answer 3

The factor $\dfrac x {1-(1-x)^\tau}$ does not change as $j$ goes from $1$ to $\tau.$ Therefore by the distributive law, it can be pulled out of the sum: $$ \sum^\tau _{j=1} \frac{(1-x)^{j-1}x}{1-(1-x)^\tau}(1-(1-x)^j) = \frac x {1-(1-x)^\tau} \sum_{j=1}^\tau (1-x)^{j-1} \Big(1-(1-x)^j)\Big). $$ Next one can apply the distributive law one more time: $$ \frac x {1-(1-x)^\tau} \sum_{j=1}^\tau (1-x)^{j-1} \Big(1-(1-x)^j)\Big) = \frac x {1-(1-x)^\tau} \sum_{j=1}^\tau\Big( (1-x)^{j-1} - (1-x)^{2j-1} \Big). $$ This becomes two separate series: $$ \frac 1 {1-(1-x)^\tau} \left( \sum_{j=1}^\tau (1-x)^j - \sum_{j=1}^n (1-x)^{2j-1} \right). $$ Each of these is a finite geometric series and can be treated accordingly.