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

Question

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.

score 0 · Answer 1 · answered Dec 12 '17 at 09:53

0

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

now try to integrate the sum.

answered Dec 12 '17 at 09:53

5xum

What is wrong? $\int\sum_{k=1}^{\infty}ka^k-1=\frac{a}{1-a}$. $(\frac{a}{1-a})'k =\frac{ak}{1-a} $. $\sum{k=1}^\infty ka^k(1-a) = a(1-a)\frac{ak}{1-a}$. And as you can see I am not getting my equality. – cutMeDown Dec 12 '17 at 10:00
@cutMeDown The line of equations you wrote makes no sense at all. You can't just take $k$ out of the summation if $k$ is the summation index! – 5xum Dec 12 '17 at 10:05
So, how should it be? – cutMeDown Dec 12 '17 at 10:08

score 0 · Accepted Answer · answered Dec 12 '17 at 09:54

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

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

$(1)-(2)$ gives

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

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

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

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

2 Answers2