On the proof that $\sum\limits_{k=0}^{n-1}\frac {a^k}{(1+a^k x) (1+ a^{k+1}x)}=\frac 1 {1-a} \left( \frac 1 {1+x} -\frac {a^n}{ 1+a^n x }\right)$ Question:-
Find the sum to $n$ terms of the following series
$$\frac{1}{(1+x) (1+ax)} + \frac{a}{(1+ax) (1+a^2 x)} + \frac{a^2}{(1+a^2 x) (1+a^3 x)} + \cdots$$

My solution:-
First of all I found out the general term of the series
$$t_n=\frac {a^n}{(1+a^n x) (1+ a^{n+1}x)} $$
Now to find out the partial fraction, I multiplied and divided the $t_n$ by $(a-1) x$. The partial fraction of the $t_n$ looks like this 
$$t_n =\frac {a^n (a-1) x }{(( a-1) x) [ (1+a^n x) (1+ a^{n+1} x)] } =\frac{( 1+ a^{n+1} x ) -( 1+ a^n x )}{((a-1)x) [(1+ a^n x) (1+ a^{n+1}x)]} =\frac 1 {(a-1) x} \left(\frac{1}{1+ a^n x} -\frac {1}{1+a^{n+1} x}\right) $$
As, $n \ge 0$,son on summing till all the $n$ terms the diagonal terms start to cancel out and we are left with
$$ S_n=\sum_{n=0}^{n-1} t_n = \sum_{n=0}^{n-1}{ \frac 1 {(a-1) x } \left( \frac 1 { 1+ a^n x } -\frac 1 { 1+ a^{n+1} x }  \right)  } = \left( \frac 1 {(a-1) x }  \right) \left( \frac 1 {1+x} -\frac 1 { 1+a^n x}  \right) $$

The answer given in the book:-
$$\left( \frac 1 {1-a}  \right) \left( \frac  1 {1+x} -\frac {a^n}{ 1+a^n x }\right)$$
Now, where did the $x$ in the $\dfrac{1}{(a-1)x}$ go and from where did the $a^{n}$ come in $\displaystyle \dfrac { { a }^{ n } }{ 1+{ a }^{ n }x }$
 A: Since the general term is $$t_j=\frac{a^j}{(1+a^jx)(1+a^{n+1}x)} \ ,\qquad  j=0,1,2,\dots,$$
we can decompose it into partial fractions as follows (so as to using finite telescoping  series). Suppose $$t_j=\frac{a^j}{(1+a^jx)(1+a^{n+1}x)}=\frac{A}{1+a^jx}+\frac{B}{1+a^{n+1}x},$$
then $$A=\frac{a^j}{1-a^{j+1}\cdot\frac{1}{a^j}}=\frac{a^j}{1-a}, \quad B=\frac{a^j}{1-a^j\cdot \frac{1}{a^{j+1}}}=\frac{a^{j+1}}{a-1},$$
and so
\begin{gather*}
t_j=\frac{1}{1-a}\left(\frac{a^j}{1+a^jx}-\frac{a^{j+1}}{1+a^{j+1}x}\right).
\end{gather*}
Therefore, 
\begin{align*}
&\sum_{j=0}^{n-1}t_j=\sum_{j=0}^{n-1} \frac{1}{1-a}\left(\frac{a^j}{1+a^jx}-\frac{a^{j+1}}{1+a^{j+1}x}\right)\\
=&\frac{1}{1-a}\left(\sum_{j=0}^{n-1}\frac{a^j}{1+a^jx}-\sum_{j=0}^{n-1}\frac{a^{j+1}}{1+a^{j+1}x}\right)\\
=&\frac{1}{1-a}\left(\sum_{j=0}^{n-1}\frac{a^j}{1+a^jx}-\sum_{j=1}^{n}\frac{a^{j}}{1+a^{j}x}\right)\\
=&\frac{1}{1-a}\left(\frac{1}{1+x}+\sum_{j=1}^{n-1}\frac{a^j}{1+a^jx}-\sum_{j=1}^{n-1}\frac{a^j}{1+a^jx}-\frac{a^{n}}{1+a^nx}\right)\\
=&\frac{1}{1-a}\left(\frac{1}{1+x}-\frac{a^{n}}{1+a^nx}\right).
\end{align*}
A: Your result is the same as the answer from the book. Proof below :
$$f(a,x)=\left( \frac 1 {1-a}  \right) \left( \frac  1 {1+x} -\frac {a^n}{ 1+a^n x }\right)$$
$$f(a,x)=\left( \frac 1 {(1-a)x}  \right) \left( \frac  x {1+x} -\frac {a^n x}{ 1+a^n x }\right)$$
$\frac  x {1+x}=1-\frac  1 {1+x}$
$\frac {a^n x}{ 1+a^n x } = 1-\frac {1}{ 1+a^n x }$
$$f(a,x)=\left( \frac 1 {(1-a)x}  \right) \left( \left(1-\frac  1 {1+x}\right) -\left(1-\frac {1}{ 1+a^n x }\right) \right)$$
$$f(a,x)=\left( \frac 1 {(1-a)x}  \right)  \left(-\frac  1 {1+x} +\frac {1}{ 1+a^n x } \right) = \left( \frac 1 {(a-1)x}  \right)  \left(\frac  1 {1+x} -\frac {1}{ 1+a^n x } \right)$$
$$\left( \frac 1 {1-a}  \right) \left( \frac  1 {1+x} -\frac {a^n}{ 1+a^n x }\right)=\left( \frac 1 {(a-1)x}  \right)  \left(\frac  1 {1+x} -\frac {1}{ 1+a^n x } \right)$$
Remark : 
$\sum_{n=0}^{n-1}$ is a bad notation because $n$ is a constant, not a variable. Better use two different symbols.
COMMENT :
Sorry, I just see the answer of Did in the comment section. It's the same proof as mine but more concise and earlier.
