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I have to calculate the partial sum for an equation. How can I calculate the sum for $$\sum_{n=1}^{\infty} \frac{1}{16n^2-8n-3}\ \text{?}$$

Thanks.

Harish Chandra Rajpoot
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Carmen Mitru
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    Partial sum to infinity ??? –  Jan 24 '16 at 11:12
  • A handful of older questions which are in a similar spirit: http://math.stackexchange.com/questions/55768/how-to-prove-sum-frac14n2-1-frac12-frac12-times-2n1 http://math.stackexchange.com/questions/265277/sum-of-this-series-sum-n-1-infty-frac14n2-1 http://math.stackexchange.com/questions/571293/sum-limits-n-0-infty-frac13-8n-16n2?rq=1 http://math.stackexchange.com/questions/306992/calculate-this-sum-sum-frac14n14n3 http://math.stackexchange.com/questions/993890/how-to-find-the-sum-of-the-series-sum-k-2-infty-frac1k2-1 http://math.stackexchange.com/questions/1412181 – Martin Sleziak Jan 25 '16 at 15:33

3 Answers3

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Notice, use partial fractions: $\frac{1}{16n^2-8n-3}=\frac{1}{4}\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right)$ as follows $$\sum_{n=1}^{\infty}\frac{1}{16n^2-8n-3}$$ $$=\frac{1}{4}\sum_{n=1}^{\infty}\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right)$$ $$=\frac{1}{4}\lim_{n\to \infty}\left(\left(\frac{1}{1}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{13}\right)+\ldots+\left(\frac{1}{4n-7}-\frac{1}{4n-3}\right)+\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right)\right)$$ $$=\frac{1}{4}\lim_{n\to \infty}\left(1-\frac{1}{4n+1}\right)$$ $$=\frac{1}{4}\left(1-0\right)=\color{red}{\frac 14}$$

Harish Chandra Rajpoot
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HINT.

This is a telescoping sum:

$$ \sum_{n=1}^{k} \frac{1}{16n^2-8n-3}={1\over4}\sum_{n=1}^{k}(a_n-a_{n+1})={1\over4}(a_1-a_{k+1}), $$ where $a_n=1/(4n-3)$.

Intelligenti pauca
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You should start with partial fraction decomposition by finding the roots of the denominator\begin{align}\frac{1}{16n^2-8n-3}&=\frac{1}{\left(4n-3\right)\left(4n+1\right)}=\frac14\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right)\end{align} In this way, the initial series can be written as telescoping series and therefore you can calculate the partial sum up to $N \in \mathbb N$ and then let $N\to+\infty$ as follows: \begin{align}\sum_{n=1}^{N}&\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right)=\\[0.2cm]&=\left(1-\frac15\right)+\left(\frac15-\frac19\right)+\left(\frac19-\frac1{15}\right)\dots+\left(\frac{1}{4N-3}-\frac1{4N+1}\right)\\[0.2cm]&=1+\left(\frac15-\frac15\right)+\left(\frac19-\frac19\right)+\dots+\left(\frac{1}{4N-3}-\frac1{4N-3}\right)-\frac1{N+1}\\[0.2cm]&=1-\frac{1}{4N+1} \end{align} So, returning to initial series \begin{align}\sum_{n=1}^{+\infty}\frac14\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right)&=\lim_{N\to+\infty}\frac14\sum_{n=1}^{N}\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right)\\[0.2cm]&=\lim_{N\to+\infty}\frac{1}{4}\left(1-\frac1{4N+1}\right)=\frac14(1-0)=\frac14\end{align}

Jimmy R.
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