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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.

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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

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    +1 for the reference, with the note that Shank's transform is significantly more general, since it only assumes $R_n \sim \alpha q^n$ ("behaves *approximately* as") while the posted question states $R_n = \alpha q^n$. – 2017-01-23
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    It's showing the technique though. I can write $S_n=S + ab^n$ since $R_n$ is given and get 3 equations for $n, n+1 $ and $ n+2$ . Solving for S will give me the result I'm looking for. – 2017-01-23
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Substitute $S_n = S-R_n = S - a b^n$ and simplify the RHS:

$$ \require{cancel} \begin{align} \frac{S_n S_{n+2}-S_{n+1}^2}{S_{n+2}+S_n-2S_{n+1}} & = \frac{(S-ab^n)(S-ab^{n+2}) - (S-ab^{n+1})^2}{\cancel{S}-ab^{n+2}+\cancel{S}-ab^{n}-2(\cancel{S}-ab^{n+1})} \\ & = \frac{(\cancel{S^2} - S a b^{n+2} - S a b ^n + \bcancel{a^2b^{2n+2}}) - (\cancel{S^2} - 2 S ab^{n+1}+\bcancel{a^2b^{2n+2}})}{2 ab^{n+1} - ab^{n+2}-ab^n} \\[5pt] & = \frac{S\,\cancel{(2 ab^{n+1} - ab^{n+2}-ab^n)}}{\cancel{2 ab^{n+1} - ab^{n+2}-ab^n}} \\[5pt] &= S \end{align} $$

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    I'm not sure if we are allowed to start from what is given as a result. Otherwise, that was easy. Thanks. If I had to figure the $S$ by myself do you know how I could do it? – 2017-01-23
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    Maybe I'm thinking of another type of exercise where I'm not allowed to do it. In my class we are told to be very careful with this because sometimes when we take exams we could lose the whole exercise because of it. But you're right, my bad. I just wondered how it could be done the other way. – 2017-01-23
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    @JohnKatsantas Your question asks: "*Prove the sum ... is given from the following*". My answer is proving just that. I am not sure I follow your "*start from what is given as a result*" reservation. When asked to prove an equality, it doesn't matter which side of it you start with, as long as you end up proving that it does indeed equal the other side, which is precisely what the above does. P.S. We crossed comment edits, which is why your reply now shows before mine. What you need be careful with is the exact statement of the problem, in particular $R_n = ab^n$ vs. $R_n \sim ab^n$. – 2017-01-23