1.54 The relations considered in this problem are used on many occasions throughout the book
(c)show also if$|\alpha|<1$,then

$\sum_{n=0}^\infty n\alpha^n = \frac{\alpha}{(1-\alpha)^2}$

• Method one – dislocation subtraction
let$X=\sum_{t=0}^n n\alpha^n = 0 + \alpha^1 + 2\alpha^2 + 3\alpha^3 + \ldots + n\alpha^n$
and$\alpha X=\sum_{t=0}^n n\alpha^{n+1} = 0 + \alpha + \alpha^2 + 2\alpha^3 + \ldots + (n-1)\alpha^n + n\alpha^{n+1}$
then we make a subtraction$(1-\alpha)X=\alpha+\alpha^2 + \alpha^3 + \ldots + \alpha^n -n\alpha^{n+1}$
let$n\rightarrow \infty$ and we will get:
$\lim_{n\to \infty}(1-\alpha)X=\lim_{n\to \infty}(\alpha+\alpha^2 + \alpha^3 + \ldots + \alpha^n -n\alpha^{n+1})\\=\lim_{n\to \infty}[(\alpha+\alpha^2 + \alpha^3 + \ldots + \alpha^n) -n\alpha^{n+1}]$

in it,$\lim_{n\to \infty}n\alpha^{n+1} \to 0$,which means that this part can be eliminated,then we will make a summation for the rest of the part,which is a form of geometric progression.
$X=\lim_{n\to \infty}\frac{1}{1-\alpha}\times(\alpha+\alpha^2 + \alpha^3 + \ldots + \alpha^n)\\=\lim_{n\to \infty}\frac{1}{1-\alpha}\times\frac{\alpha(1-\alpha^{n+1})}{1-\alpha}\\=\frac{\alpha}{(1-\alpha)^2}\lim_{n\to \infty}(1-\alpha^{n+1})\\=\frac{\alpha}{(1-\alpha)^2}$
• Method two
Note that if we expand this complex summation formula and combine some parts,perhaps it will be a little easy.
$\sum_{n=0}^\infty n\alpha^n = \lim_{n \to \infty}(0+\alpha^1+2\alpha^2+3\alpha^3+\ldots+n\alpha^n)\\=\lim_{n\to \infty}[(\alpha+\alpha^2+\alpha^3+\alpha^4+\ldots+\alpha^n)\\+(0+\alpha^2+\alpha^3+\alpha^4+\ldots+\alpha^n)\\+(0+0+\alpha^3+\alpha^4+\ldots+\alpha^n)\\+(0+0+0+\alpha^4+\ldots+\alpha^n)\\+(0+0+0+0+\ldots+\alpha^n)]\\=\lim_{n\to \infty}[\frac{\alpha(1-\alpha^n)}{1-\alpha}+\frac{\alpha^2(1-\alpha^{n-1})}{1-\alpha}+\ldots+\frac{\alpha^n(1-\alpha)}{1-\alpha}]\\=\lim_{n\to \infty}\sum_{k=1}^n \frac{\alpha^k(1-\alpha^{n+1-k})}{1-\alpha}\\=\lim_{n\to \infty}\sum_{k=1}^n (\frac{\alpha^k}{1-\alpha}-\alpha^n)\\=\lim_{n\to \infty} \frac{\alpha(1-\alpha^k)}{(1-\alpha)^2} -\lim_{n \to \infty} n\times\alpha^n$

There are two parts,in which,$\lim_{n \to \infty} n\times\alpha^n \to 0 (|\alpha|<1)$,as a result
$\sum_{n=0}^\infty n\alpha^n =\lim_{n\to \infty} \frac{\alpha(1-\alpha^k)}{(1-\alpha)^2}= \frac{\alpha}{(1-\alpha)^2}$
• Method three,as the manual solution to this book gives:
Note that$\sum_{n=0}^\infty \alpha^n = \frac{1}{(1-\alpha)}$
then differentiating both sides of it,we get
$\frac{d}{d\alpha}(\sum_{n=0}^\infty \alpha^n) = \frac{d}{d\alpha}(\frac{1}{1-\alpha})$
finally$\sum_{n=0}^\infty n\alpha^n = \frac{\alpha}{(1-\alpha)^2}$