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Assume the $n$-th partial sum of a series $\sum_{n=1}^\infty a_n$ is the following: $$S_n=\frac{8n-6}{4n+6}.$$ Find $a_n$ for $n > 1$.

I'm really stuck on what to do here.

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  • $\begingroup$ Do you understand that you need to find a sequence $(a_n)_{n\in \mathbb N}$ such that $$\forall n\in \mathbb N\left(\sum \limits_{k=1}^n(a_k)=\dfrac{8n-6}{4n+6}\right)?$$ $\endgroup$ – Git Gud Mar 22 '15 at 0:48
  • $\begingroup$ S(n)-S(n-1) should be the nth term. So S(2)-S(1) should be second term and so on $\endgroup$ – Valtteri Mar 22 '15 at 0:48
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Hint. Note $a_n = S_n - S_{n-1}$ for $n > 1$.

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Observe that for each $n\in \mathbb{N}$,

$a_{n+1}=S_{n+1}-S_n= \dfrac{8(n+1)-6}{4(n+1)+6}-\dfrac{8n-6}{4n+6}$ and $a_1=S_1=\dfrac{8.1-6}{4.1+6}=\dfrac{1}{5}$.

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    $\begingroup$ Continuing a bit: $$\frac{8(n+1)-6}{4(n+1)+6}-\frac{8n-6}{4n+6}=\frac{8n+2}{4n+10}-\frac{8n-6}{4n+6}$$. Put over a common denominator gives: $$\frac{(8n+2)(4n+6)-(8n-6)(4n+10)}{(4n+10)(4n+6)}$$ Simplify: $$\frac{3n^2+56n+12-32n^2-(80-24)n+60}{16n^2+64n+60}$$ $$\frac{3n^2+56n+12-32n^2-56n+60}{16n^2+64n+60}$$ $$\frac{72}{16n^2+64n+60}$$ $\endgroup$ – FundThmCalculus Mar 22 '15 at 1:42

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