Prove the sum of series is given from the following if the remainder $R_n=ab^n$ Prove the sum of a series $\sum_{k=0}^\infty c_k$ is given from the following  
$$ S=\frac{S_n S_{n+2}-S_{n+1}^2}{S_{n+2}+S_n-2S_{n+1} }$$
 if the remainder can be written as : $R_n=ab^n$ 
It is given that $a,b$ are constants and the series converges.
All I know is $$S-S_n=R_n$$ and that the $\lim_{n\to \infty}R_n=0$ since the series converges and maybe $b<1$ so the limit can be $0$. I don't know have any other idea though.
 A: Substitute $S_n = S-R_n = S - a b^n$ and simplify the RHS:
$$
\require{cancel}
\begin{align}
\frac{S_n S_{n+2}-S_{n+1}^2}{S_{n+2}+S_n-2S_{n+1}} & = \frac{(S-ab^n)(S-ab^{n+2}) - (S-ab^{n+1})^2}{\cancel{S}-ab^{n+2}+\cancel{S}-ab^{n}-2(\cancel{S}-ab^{n+1})} \\
 & = \frac{(\cancel{S^2} - S a b^{n+2} - S a b ^n + \bcancel{a^2b^{2n+2}}) - (\cancel{S^2} - 2 S ab^{n+1}+\bcancel{a^2b^{2n+2}})}{2 ab^{n+1} - ab^{n+2}-ab^n} \\[5pt]
 & = \frac{S\,\cancel{(2 ab^{n+1} - ab^{n+2}-ab^n)}}{\cancel{2 ab^{n+1} - ab^{n+2}-ab^n}} \\[5pt]
 &= S
\end{align}
$$
A: Preface: Note that your condition $R_n = ab^n$ implies that the result can be written in the form $S_n = S + ab^n$ (where $S$ is the limit of the partial sums and $S_n$ is the $n$th partial sum) which Wikipedia writes as $A_n = A + \alpha q^n$ 
Wikipedia has a page on this. It is called the Shank's Transformation, and it's purpose is to increase the rate of convergence of a given series by transforming the series into something new. Here is the passage you need from that page:

Motivation: 
The Shanks transformation is motivated by the observation
  that — for larger  $n$ — the partial sum  $A_n$ quite often behaves
  approximately as $A_n = A + αq^n$, with  $|q|<1$   so that the
  sequence converges transiently to the series result  $A$ for $n \to
 \infty$. So for $n − 1, n,$ and  $n + 1$  the respective partial sums
  are: $A_{n − 1} = A + α q^{n − 1} , A_n = A + α q^n,$ and $A_{n + 1} = A +
 α q^{n + 1}$ .
These three equations contain three unknowns:  $A, \alpha ,$ and $q$.
  Solving for  $A$ gives $A = \frac{A_{n + 1} A_{n − 1} − A_n^2}{A_{n +
 1} − 2 A_n + A_{n − 1}}$ . 
In the (exceptional) case that the denominator is equal to zero: then 
  $A_n = A$ for all $n$

Note that to get your form of Shanks Transformation, replace all $n$ by $n+1$ in the formula from Wikipedia
