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I have to show, that this series converges and determine its limit.

$$\sum\limits_{n=0}^{\infty}\frac{1}{3-8n-16n^2}$$

So far, my idea was to transform it into a telescoping series, but I don't really know how to do it.

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  • $\begingroup$ Have you factorised it? $\endgroup$ – user85798 Nov 18 '13 at 2:07
  • $\begingroup$ What techniques have you studied to find the sum of a series? $\endgroup$ – Mhenni Benghorbal Nov 18 '13 at 2:17
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Factorise and use partial fractions. $\dfrac{1}{3-8n-16n^2}=\dfrac{1}{4}\left(\dfrac{1}{4n+3} - \dfrac{1}{4n-1}\right)=\dfrac{1}{4}\left(\dfrac{1}{4(n+1)-1} - \dfrac{1}{4n-1}\right)$

So $\displaystyle\sum\limits_{n=0}^{\infty}\frac{1}{3-8n-16n^2}=\frac{1}{4}$

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HINT: Use the factorization $3-8n-16n^2=(3+4n)(1-4n)$ to split the fraction into partial fractions:

$$\frac1{3-8n-16n^2}=\frac{A}{1-4n}+\frac{B}{3+4n}=\frac{B}{4n+3}-\frac{A}{4n-1}$$

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We have \begin{align} \dfrac1{3-8n-16n^2} & = \dfrac1{(3+4n)(1-4n)} = \dfrac14 \cdot \dfrac1{4n+3} - \dfrac14 \cdot \dfrac1{4n-1}\\ & = \dfrac14 \cdot \dfrac1{4(n+1)-1} - \dfrac14\cdot \dfrac1{4n-1} \end{align} Hence, $$\sum_{n=0}^N \dfrac1{3-8n-16n^2} = \dfrac14 \cdot \left(\sum_{n=0}^N \dfrac1{4(n+1)-1} - \dfrac1{4n-1} \right) = \dfrac14 \cdot \left(\dfrac1{4(N+1)-1} + 1\right)$$ Now let $N \to \infty$ to get the limit.

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