Sum up the series: $sum_{k=1}^{infty}ka^k(1-a) = frac{a}{1-a}$

How can I show that

$\sum_{k=1}^{\infty}ka^k(1-a) = \frac{a}{1-a}$

I tried to integrate the series, but that did not help me.

Answer

$S = a(1-a) + 2a^2(1-a) + 3a^3(1-a)+\cdots + \infty \tag1$

$aS = a^2(1-a) + 2a^3(1-a) + 3a^4(1-a) +\cdots +\infty \tag 2$

$(1)-(2)$ gives

$$(1-a)S = a(1-a) + a^2(1-a) + a^3(1-a) + \cdots+\infty$$

$$(1-a)S = a(1-a) \left[1+a+a^2+a^3+\cdots\right]$$

$$(1-a)S = a(1-a).\frac{1}{1-a} = a$$

$$S = \frac{a}{1-a}$$